ZBIRKA ZADATAKA IZ
MATEMATIKE
za II razred sredwe {kole
M6 M7 M8
M9 M10 M11
M4 M11
6 M7 M8
9 M10 M11
M4 M5
M
M
f (x)= xn n∈N
= ax 2 + bx + c y = mx + n
y
α α = 29◦ 18 + 62◦ 17
i =25◦ 38 r =18◦ 7
l =0, 993563 0, 002536 2ϕ
g =9, 806059 0, 025028 2ϕ ϕ l g ϕ =38◦ 16 30
α =15◦ 17 c =75 a =5 β =73◦ b =56 c =65 a =360 b =599
a =23 α =19◦ 37
a =35, 208 β =72◦ 31
c =50 = α =68◦ 27
c =112, 28 β =59◦ 8
c =44 a =17, 3
c =15, 214 b =8, 735
a =83 b =51
a =7, 24 b =11, 63
α =16◦ 40 b =600 α =68◦
n ∈ N a ∈ R\{0} a n = 1 an a 0 =1. ab ∈ R\{0} mn ∈ Z
1◦ am · an = am+n
2◦ (am )n = anm
3◦ am : an = am n
4◦ (ab)n = an · bn
5◦ Å a b ãn = an bn
1◦ a x ∈ R+ ∪{0} n ∈ N n √a = x ⇔ xn = a
2◦ a 0 b 0 n ∈ N n √ab = n √a · n √b
3◦ a 0 b> 0 n ∈ N n … a b = n √a n √b
4◦ a 0 m,n ∈ N m » n √a = mn√a
5◦ a 0 m,n,p ∈ N np√amp = n √am
(( 1) 1 ) 1 > 0 (( 1) 1 ( 1) 1 ) 1 > 0 ( 1)2n +(+1)2n (+1)2n+1 ( 1)2n+1 ,n ∈ N 9 3 4 12 ( 6) 1
(a 3 ) 2 · Ç b3 a3 å 3 a =0 b =0
(a 5 b7 )3 Ç ab10 c6 å 3 a =0 b =0 c =0 n ∈ N) ( an )4n :( an )2n+1 (( a)n )2n :( a 3 )2n+1
a> 0 n ∈ N x> 0 x< 0 x =(a)n x =( a)2n x =( a)2n+1 x =( a)2n+3
a< 0
0
2 >a a 2 = a a 2 <a an+1 >an n ∈ N an+1 <an n ∈ N 7 4 76 124 12 3 1014 10 8 10 4 316 :318 9 5 :9 7 (4 5 ) 1 (62 ) 2 63 5 4 (5 2 ) 4
· xn · x 1 n axn ·
n 1 · bx 2xn y 2 · 5x 1 n y n 1 (x +2y )1+n · (x +2y )1 n · (x +2y )n 2 ax (a + b)2x y · ay (a + b)y x (a b)9 · (b a)6 (a x)n · (x a)2n xm (xn + x) xn (x + xm ) (3
4a
+5ax )( 2ax y ) (x 2n xn y m + y 2m )(xn + y m ) (a 2n + an +1)(a 2n an +1) 5x +5x+1 +5x+2 xn+3 2xn+2 + xn+1 (x +3y )4n · (x 3y )4n Ç 9x2 1 4y 2 1 ån Å 2y +1 3x +1 ãn Å 2y 1 3x 1 ãn (a 2 +2a +4)m · (a 2)m+n · (a +2)n =(a 3 8)m · (a 2 4)n
63a 3n x bn :9a 2n x bn 1
48xa+b y a b :( 16xa b y b a )
24an (1 a)n 1 :12an 1 (1 a)n 2
(x + y )m+n (x + y )m n (x y + z )m 2n (x y + z )2m 3n
28a2n b1 n 33xa y 2 b · 22x2a y 1 b 49an b2 n 3an+1 bn 1 4x2 m y 2+m : 3an 1 b1 n 5x1 m x1+m
6x +6x 1 +6x 2 +6x 3 x 2a+b + xa+2b + xa+b+2 x 2a b + xa + xa b+2
Äxn+2 2xn+1 + xn ä : xn Ä7a 2 9ab +2b2 äm+n :(7a 2)m+n
Å a + b x y ã4 Å 1 a + b ã2 : Å a + b x y ã3
Å x + y a b ãn+1 : Å x + y a b ãn : x2 y 2 a2 b2
36((a 3 )x )y :9((a 2 )y )x ((36a 3 )x )y :((9a 2 )y )x Ç 5x5 4y 4 å2 m : ÇÇ x5 8y 4 åm å2
Ä9x 2 49y 2 äx y x+y :((3x +7y )x+y )x y xm2 n2 (x + y )a2 +2a (x y )a3 +b3 x 8 · x 10 · x 6 3x 5 y 3 · 5x 3 y 8 b3 y 2 c 3 z 3 · 3b 3 y 5 5c5 z 4 · 5by 2 6c 1 z 2
(R, +, ) x n x =0,n ∈ N xn x n xn m n
2m · 3n 1 2m 1 · 3n 2m · 3n m ∈ Z n ∈ Z
5n+1 2n 2 +5n 2 2n 1 10n 2 n ∈ Z x5 + x12 x 5 + x 12 y 29 y 21 y 21 y 29 a5 + a6 + a7 a 5 + a 6 + a 7
3 + b2 + b +1
3 + b 2 + b 1 +1 45n+1 32n+1 · 5n 60n 22n · 3n 1 · 5n+1 n ∈ Z
2n +2n =2n+1 21 n 2 n =2 n 2 · 3n +3n =3n+1 2 n +2 n+1 =3 · 2 n 3n+1 3n 2 2n +2 n 4n +1 (3n +3n 1 )2 9n 1
n
6
)
f (x)= x n n∈N
f (x)= xn n∈N
f (x)= x 14
f (3) f ( 3) f (5) f (0) f ( 8) f (0) f ( 6) f (8) f ( 3) f ( 1) f (4) f (7) h(x)= x 27
h(4) h( 4) h( 10) h(0) h(12) h(0) h( 82) h(45) h( 5) ( 2) h(7) h(9)
A(2;256) B ( 2;256) C ( 3; 6561) y = x 9
A(2;512) B ( 2; 512) C ( 3;19683) y = x 8 78 98 0, 48
)=
)=121
)=80
x 4 3 : x 2 3 10a 3 4 :5a 1 4 4 3 x 1 2 y 5 6 :4x 1 2 y 1 3
Å5a 5 6 6a 3 4 ã : Å 2a 1 12 ã Å12x 3 5 10a 1 5 ã : Å 4a 1 10 ã
Åx 3 4 ã 2 7 Åa 5 3 ã 3 10 Åy 5 2 ã 3 Åa 3 4 · b 1 3 ã 4 Ñ 16a 1 6 b 4 3 é 3 4 Ñx 15 10 Åx 3 10 x 0,5 ã 1 2 é 3 2 R+ 4 x 1 2 1 6 · Ñ x 3 4 8 é 1 9 Ñ27 a 1 2 b0,2 é2,5 · Ñ b 1 12 3 4 √3 · a 1 24 é1,2 a ∈ R+ a 8 a 60 a 21 a 16 a 7 a 3 a a 0,25 a 1 a 1 3 x ∈ R+ x 6 x 18 x 5 x x 0,5 x 3 2 x 1 3 x 0,2 x 5 9 2 1 2 ≈ 1, 41 2 3 2 2 5 2 2 1 2 2 5 2 x ∈ R+ y ∈ R+ x y y = x 2 3 y = x 5 7 y = x 4 3 y = x 0,75 y =9x 4 5 y = 1 4 x 2 3 x 1 2 y 1 2 Åx 1 2 + y 1 2 ã c 1 3 d 1 3 Åc 1 3 d 1 3 ã Åx 2 3 1ãÅx 1 3 +2ã Åa 3 4 +2ãÅa 1 4 3ã
Åa 1 3 b 1 3 ã2 ÅÅx 1 4 y 1 4 ãÅx 1 4 + y 1 4 ãã2
9n 9n 1 1 2 (27n 1 19 · 27n 2 ) 1 3 8n 2 +7 8n 3 1 3 (16n 1 16n 2 ) 1 4
(a,b) a b (a,b) (c,d) (a + c,b + d) (a,b)+(c,d)=(a + c,b + d). (a,b) (c,d) (ac bd,ad + bc) (a,b) · (c,d)=(ac bd,ad + bc). (0, 1) i (a,b) (a,b)= a + ib.
(a + ib)+(c + id)=(a + c)+ i(b + d), (a + ib) (c + id)=(ac bd)+ i(ad + bc), a + ib c + id = ac + bd c2 + d2 + bc ad c2 + d2 i,c2 + d2 =0. a 2 + b2 =(a + ib) · (a ib) i2 = 1 ax 2 + bx + c =0,a,b,c ∈ C
. ax 2 + bx + c = a(x x1 ) (x x2 )
f (x)= ax 2 + bx + c,a =0,a,b,c ∈ R, f (x)= a Åx + b 2a ã2 + 4ac b2 4a a> 0 x = b 2a y = 4ac b2 4a a< 0 x = b 2a y = 4ac b2 4a x 2 +9=0 3i 3i x 2 +3=0 √3i √3i x 2 +8=0 2√2i 2√2i 6 7i 2 (√3 1)i 2 1 7 m 3+7i m m n
(3n, 8) (9, 4m) √3+2i 1 n 1 m i (m +2)+(n 4)i
(2, 1)3 ( 2, 1)3 i21 i17 + i36 + i42 i135 + i235 (6, 2)+(4, 3) ( 3, 2) ( 3, 2) (4√7 6√8i) ( √28+2i) (2 i)+ Å 4 3 7 2 iã (2, √3) ( 1, 7) 1 2 1 3 i 1 √2i
(0, 3) (7, 4) ( 2, 5) (1, 4) (1 i)(2 3i) (1+ i)2 (7 5i)3 6 i 3+5i (2 i) (3+4i) (1 i)3 1
2 +4 9x
+16
3x 2 +1 (1+ i)4 (1 i)4 i + i2 + i3 + i4 + ··· + i2000 z = Å 1+ i 1 i ãn n ∈ N x 2 5x +6 2x 2 7x 3 x 2 2x 7 3x 2 + x +1 2x 2 ix +2 i x
x 2 5x +6 x 3 2x 2 +5x 7 2x +1
a b (a + b)x 2 +(a b)x +3 x 2 + x +3 x 2 9x +8 x 2 +7x 60 2x 2 5x +6 x 2 x 2 7x +9 x +2 ax 2 + bx + c x 2a a + b c x 2 2x =0 x 2 +3x =0 (x 7)(x +3)+(x 1)(x +5)=102 5x2 +9 6 4x2 9 5 =3 x x +1 + x x 1 =2 2 3 x = 1,x =1 13x 2 +12x 1=0 x 2 +12x 64=0 x 2 4x =45 x 2 +14x +24=0 x 2 11x 60=0 x 2 4 1 2 x +4, 5=0 x 2 +3 5 12 x +2=0 2x 2 7x +6=0 5x 2 8x +3=0 (x 3)2 +(x +4)2 (x 5)2 =17x +24 (x +5)2 +(x 2)2 +(x 7)(x +7)=11x +30 x x 2 + 2 (x 2)(x 3) = 3 x 3
x
x
x 2 10x +9 x 2 4x 60 x 2 +25x +114
x 3y =1
x 2 y 2 =8, y +3=4. 2x 2 y =2, x y =1; xy =15, 2x y =7; x 2 + y 2 =40 x + y =8 x 2 + y 2 =6y y +2x =0 (x 2)(y +1) 1=0 x y 3=0. 3 x +5 + 2 y 3 =2 4 x 2 1 y 6 =0 3x 2 y +5 + y x =2 x y =4. x 2 +2xy + y 2 x y =6, x 2y =3 2x 2 +5x 2 3=0 x 4 5x 2 +4=0 x 4 9x 2 +10=0 x 4 5x 2 +6=0 x 2 (m +1)x + m =0 m m x 3 =1 x 3 = 1 3x 2 +2=0 x 2 + y 2 =5, 4x 2 +9y 2 =25 3x 2 +5y 2 =323, 7x 2 12y 2 =375.
x 2 + y 2 =25, 2x 2 +9y 2 =162 x2 64 + y 2 12 =1, x 2 + y 2 =25. x 2 +4y 2 =100 x 2 2y 2 =4 x 2 y 2 3 =1, x2 16 + y 2 12 =1. 3x 2 +7x +4=0 2x 2 3x 2=0 3x 2 + x 4=0 x 2 20x +64=0 x 2 6x +9=0 x 2 4x +5=0 k 2x 2 3x +(k 1)=0 k =0 kx2 +2(k 6)x + k 3=0 k x 2 (k +1)x +(2k 1)=0 p1 ,p2 ,q1 ,q2 p1 p2 =2(q1 + q2 ) x 2 + p1 x + q1 =0 x 2 + p2 x + q2 =0
f (x)= ax 2 + bx + c a b c
f (2)=0 f (3)= 7 f ( 2)=8
f ( 6)= 3 f (1)=5 f (2)=25
f (5)=0 f (4)= 3 f (6)=5 y = x 2 + px + q p q x =1 y = 4 y = x 2 + px + q p q x = 5 2 y = 9 4
y =(k +2)x 2 +(1 k )x + k, (k = 2) k x =2 y = x 2 4x +3 y = x 2 +2x +3 y =2x 2 x +1 y =(x 1)(x +2) 3 y =(x 2)2 +5 y = x 2 3x 4
y = x 2 4 y =2x 2 + x 1 y = 2x 2 x +6 y =(x 3)2 1 y =(x 1)(x 2) (x 3)(4 x) y = 2x 2 5x 6 c< 0 a + b + c =0 ax 2 + bx + c =0 y = x 2 −|x|
x 2 + x 2 > 0 5(x 2)2 > 0 2x 2 +3x 2 0 4x 2 +3x 1 < 0 x 2 +3x +3 > 0 x 2 x 1 > 0 ∗ 4(x 3)(x +2)(x +5) > 0 ∗ 5(x +2)2 (x 3) > 0 x 1 x +2 > 0 3x +7 2 5x > 1 x 1 2x +1 > 1 k x x 2 2x + k> 3 x 2 2x + k< 3 k x 2 +(k 2)x +1=0 k x 2 + kx +1=0 k x (k 1)x 2 +2kx +3k 2 > 0 y = ax 2 + bx + c y = mx + n y = x 2 10x +24, y = x +6 y = x 2 + x +1, y = 1 x 2 y =14, y x = 2 y x 2 =0, 5x y =6 x 2 + y =2x 1, x + y = 1. x 2 3x +2=0 x 2 8x +7=0 x 2 + x +1=0 2x 2 +5x +3=0 x 2 4=0 1 (x 1)(x 2)=0
k (2k x)(2k + x)= kx2 +5 (k 1)x 2 +2k =8 k (k 1)x 2 2(k +1)x + k 2=0 k 2x 2 +(2k 1)x + k 1=0 3x1 4x2 =11 S n S = 1 2 n(n +1) x1 x2 3x 2 +2x 5=0 x 2 1 x2 + x1 x 2 2 x 3 1 + x 3 2 x1 x2 2x 2 4x +8=0 (x 2 16x)2 2(x 2 16x) 63=0 x 2 16x = n x 2 (11 7i)x +18 39i =0 n 1,n,n +1 a =16 b =18 c =20
2 y = 2x 2 +3x +17 y = x 2 +5x +6 y =(x 1)(x 2)+(x 3)(x 4) y = 2 5 x 2 y = 2 5 x 2 +3 y = 2 5 x 2 +4x +3 y = (x +2)2 y = x 2 +(x 1)(x +2) y = 2x 2 +3x +4 b y = x 2 + bx +7 x =4
b c y = ax 2 + bx + c ( 1, 1), (2, 1) y = 5x 2 +6x 7 y = x(x 1) y = x 2 y = x 2 (x 1)(2 x) x 2 > 0 x 2 (x 1)(2 x) 0 3x 2 2x 0 14 5x x 2 < 0 x 2 3x +4=0 2x 2 +6x 7=0 1 2 x 2 +3x +4=0 x 2 4=0
∈ R a + b + c =5 ab + ac + bc =8
1, 7 3 ã
m x 2 (m +3)x + m +2=0 x1 x2 1 x1 + 1 x2 > 1 2 x 2 1 x 2 2 < 5 (m +1)x 2 (m +2)x (3 2m)=0 m x1 x2 x 2 1 + x 2 2 x1 + x2 = 7 9 x1 x2 x1 + x2 +2x1 x2 =0 m(x1 + x2 ) x1 x2 =3m +4 m m (m 2 m 2)x 2 +(2m 2 2m +5)x + m 2 m 2=0 m m =2; 1 m m ∈ R x 2 2(m 3)x +11 5m =0 m ∈ R 1 x1 + 1 x2 =1 |x 2 +2x|−|3 x| = x 2 x 2 + y 2 =25, x 2 +2y 2 =41. x 2 +4y 2 =20, 3x 2 y 2 =47. x 2 x + c =0 α β c x 2 + αx β =0
a b c ax 2 + bx + c =0
◦ l 3x 2 +17x 14=0 3x 2 1 +5x1 x2 +3x 2 2 4x1 x 2 2 +4x 2 1 x2 x1 x2 k kx2 (1 2k )x + k 2=0 p q x 2 + px + q =0 p q k (k ∈ R) x 2 (2k +1)x + k 2 +2=0 x2 = 1 2 x1 x1 x2
p 3i (x 2 + x +1)(x 2 + x +2) 12=0
(x 4)(x 5)(x 6)(x 7)=1680 ax 4 + bx2 + c =0
x + xy + y =11, x 2 y + xy 2 =30.
x 2 xy y 2 = 11, (x 2 y 2 )xy =180. x(x +1)(3x 2 +5y )=144, 4x 2 + x +5y =24. x 2 3xy +2y 2 =0 x 2 3x y +3=0. x 3 +1=0 x 3 8=0 x 4 1=0
x 2 +5x +4 5 x2 +5x +28=0 k x 2 2(3+ i)x + k =0 ax 2 + bx + c =0 a b c α α = p q p q q =0 p|c q |a
a> 0 x,y ∈ R ax · ay = ax+y . a> 0 x,y ∈ R (ax )y = ax y .
a,b> 0 x ∈ R ax · by =(a · b)x y = ax a> 0,a =0 f : A → B
2π 5√2 Å 1 2 ã√3
b a b > 0 a> 0 b a b > 1 b a b < 1
<a< 1
b > 1
f1 (x)= x 3 f2 (x)= π 3x f3 (x)=3 x f4 (x)= m0 e λt
f5 (x)= xx f6 (x)=1x f7 (x)= xy f8 (x)= y x
f (x)=1, 5x g (x)=3x f (x)=0, 3x g (x)=0, 9x
f (x)= Å 1 6 ãx g (x)=6x
f (n)=2n n ∈ N f (10) f ( 30) f (f (0)) f ( 3x)
f (x)=100x f (x)=2 x f (x)=2x +3x
f (x)=0, 1x 0 x 1
f (x)=3x
f1 (x)=3x +1 f2 (x)= 1 2 · 3x f3 (x)= 3x
f (x)=2x g (x)=2x+3 f (x)=0, 3x g (x)=0, 3x 1
f (x)
f (x)=2x 8 f (x)=5 x 0, 04
f (x)=(0, 5)x 5 g (x)=2+2x h(x)=1 2 3x y
f (x)
f (x)=17x+5 f (x)=5 · 7x 1 +6 f (x)=3x +5x
f (x)
f (x)=25x +25 f (x)= Ç √3 2 å x
f (x)
f (x)=2x 1 f (x)= 2x +3
f (x)=3x−|x| f (x)= Å 1 2 ã x2 |x| Å 5 6 ã 2 3 > 1 Å 2 3 ã 3 4 < 1 (0, 15) 0,81 > 1 x 0, 1x =11 Å 3 5 ãx =7 π x =7
f (x)= ® 3x x< 0 3x +1 x 0 f (x)= ® x2+1 x 0 1, 5x x> 0.
f (x)= ® x 2 x 2 x< 0 2x 2 x 0.
a 2 + b2 =7ab
a + b 3 = 1 2 ( a + b) a b ( b a) b a
a 2 + b2 = c 2
b+c a + c b a =2 · c+b a · c b a
f = {(1, 2), (2, 3), (3, 5), (4, 4)} f = {(1, 1), (2, 2), (3, 2)}
A = {−1, 0, 2},B = {0, 1, 2},f : A → B,f (x)= |x|
f 1
f (x)= 1 2 x f (x)= x +1
f (x)= 3x +17 f (x)= 5 7x
f (x) f 1 (x)
f (x)=2x +3 f (x)=5 x
f (x)= x 1 f (x)=3x 2,x 1
f (x)= x 2 +1
f (x)= x 2 x 0
f (x)=5x f (x)=2x +1 f (x)= x
f (x)= x f (x)= x +2
f 1 (f (x))= x x f
f (x)=2x 3 f (x)= x f (x)= √x
f (x)= {(1, 1), (2, 3), (3, 4)}
f (x)=1, 5x f (x)=0, 2x f (x)= Å 4 5 ãx
f (x)=2, 5x f (x)=7x
f (x)= 3 x; g (x)= 3,5 x; h(x)= 4 x
f (x)= 0,2 x; g (x)= 0,3 x; h(x)= 0,5 x
f (x)= a (5x 3) f (x)= a ((3x 5) · (2 x))
f (x)= a x 2 1 x f (x)= a (x 2 +1)
f (x)=5+ 3 x f (x)= 0,2 x
f (x)= 2 (x 3) f (x)=1 2 2 x f 1 (x)
f (x)=2 · 3 (x +4) f (x)= 1 3 (2x 1) x f (x)
f (x)= 3 5 (2x +5) f (x)= 4 (1 x)
f (x)= 2 |x| f (x)= 1 2 |x|
f (x)= | 2 x| f (x)= | 1 2 ( x)|
f (x)= ® 1 x,x< 1 2 x,x 1 f (x)= ® x 2 8,x< 3 3 x,x 3
m = m0 e λt m0 λ 0, 693 T12 T12 t m 2 (3x 2)= 0,5 x (1 x) 7 x =2 (x 3) x xx 28= 27x x (100x)1+ x =(10x)3 x x x =106 · x x x +10x x 11=0 x> 0 1+ a x( a (x 1) > 0 1 a x + 1 1 a x > 0(a> 1) 8 (x 2 4x +3) < 1 1 2 x 6 · x 1 2 > 1
M = 1 2 ph
h P = M + B + B1
1 M M = p + p1 2 h p,p1 h V = 1 3 H (B + BB1 + B1 ).
M =2πRH
P =2πRH +2πR2
V = πR2 H. M = πRl l R
P = πRl + πR2 = πR(l + R)
V = 1 3 πR2 H
M = π (R + r )l R r l P = πR2 + πr 2 + π (R + r )l V = 1 3 πH (R2 + Rr + r 2 )
P =4πR2
P =2πRH
V = 4 3 πR3 V = 1 3 πH 2 (3R H ) V = 2 3 πR2 H
=25
1 =15
=6370
=4 b =6 c =12
(180◦
(180◦
(180◦ + α)=
(180◦
)=
(180◦ + α)= α, (180◦ + α)=
(360◦
(360
= a · f (x)
= f (x c)
0
= f (x)
= f (x) c c> 0 c
= f (x) x c
=2 x y =(1/2) x y = 2x y = (x/3)
=3 x y =3/4 x y = (2x/3) y = 4x
= (x π /3) y = (x + π /2) y = (x/3+ π /2)
= (3x π /2) y = (2x/3 π /4) y = (x/2+ π /3)
=2 x y =(1/2) x y = 2x y = (x/3)
= 2 x 3/4 x y = (2x/3) y = 2x
(x π /3) y = (x + π /4)
y = (x + π /2) (x π )
y = (x π /2) y = (x + π /4)
y = (2x π /2)
y = (3x + π )
y =( 3/2) (2x π /3) ( 1/2) (2x π ) ( 5/2) ((x/2)+ π /2)
<α< (π /2)
α =0, 6 α α =1/4
=3/5
= 0, 6
=40/41
=1/3
α =3/5 0 <α<π /2
α =8/17 α
α = √3/3 π /2 <α<π
α = 2, 5 α β
=9/41, β =40/41
(α + β )+ (α β )=2
(α + β ) (α β )=2
(α β ) (α + β )=2 α · β
(α + β ) · (α β )= 2 α 2 β (α + β ) (α β )= 2 α 2
+
)
)+2
+ β )+
)
)
=1/2 β =1/3 (α + β ) (α β ) ( α + β )/( α β )= (α + β )/ (α β )
)/(
)/(2
) (2 α + 2α)/(2 α 2α)
(1 2α + 2α)/(1+ 2α + 2α)
(2π /5)+ (π /5)
(11π /12)+ (3π /4)
(π /6+ α) (π /6 α) (α π /3)+ (α + π /3)
x = 1 x =0 x = 1 x = √3/3 x = √3 x = √3 x = √3/3
2x =0 (3x/2)= 1 (x π /6)=1
(x +30◦ )= √2/2 (2x 45◦ )=1/2
(π /4 x)= √3/2 (3x +2π /3)= 1/2
3x =1 (2x/3)=0 (x + π /3)= 1
(60◦ 2x)=1/2 (x/2+ π /4)= √3/2
(2π /3 x)= 1/2 (x/3 45◦ )= √2/2
2x = 1 (x/3)=1 (π /2 x)=0 (x/2 30◦ )= √3 (3π /4+ x)= 1 (120◦ +2x)= √3/3 (3x/2 π /3)= √3
(x/2)=1 4x =0 (2x/3)= 1 (2x π /3)=1 (x +120◦ )= √3/3
(5π /6 x/2)= √3 (135◦ +3x)= 1 2 x 1/2=0 1 4 2 x =0 4 2 x =3 2 x 1=0 3 2 x =1 2 x 3=0
2x · (x π /4)=0
(2x 120◦ ) (x +60◦ )=0
(x/2) · (2x π /2)=0 (x/3) · (3x + π /2)=0 x · x =1 2 x 2 x =2 x · x =3/2
a α = b β = c γ
c 2 = a 2 + b2 2ab γ
a 2 = b2 + c 2 2bc α
b2 = c 2 + a 2 2ca β
a =8 b =3 γ =60◦
a =8 b =3 c =7
a =37 β =86◦ 30 γ =50◦ 50
a =450 α =87◦ 50 β =10◦ 50
a =87 b =65 α =75◦ b =360 c =309 γ =21◦ 30
CA =1447, 4 CB =3225, 8 ACB =57◦ 12
CA =290 CB =600 ACB =100◦ 19
AB =10 AC =8
AB =12 AC =8 BAC =45◦ AC =6 BC =8 BCA =60◦
( 9◦ 31 + 21◦ 43 ) (9◦ 31+21◦ 43 )=(0, 16533+0, 37002) 31◦ 14 =0, 53535 0, 51852=0, 01683
α =0, 48938+0, 46510 α =0, 95448 α =43◦ 40
η = 0, 43261 0, 31095 η ≈ 1, 39
l =0, 993563 0, 002536 0, 23260 ≈ 0, 9930 g ≈ 9, 6984
=70◦ 23 β = b a b = a β b =23 70◦ 23 =23
28 55 0, 00324 1 12
a 48 a 32 x 2n 2 y 6n 2 x 18
32 30 3 2 02 3 2 30 ( 3)2
04 30 03 1 40 016 0 20 00 1 9 2 1 250 13 36 8 81 5 9 16 a3 b9 a12 c18 b9 a 2n 2 n a 2n 2 6n 1
x> 0 x> 0 x< 0 x< 0
x< 0 n x> 0 n x> 0 x> 0 x> 0 a> 1 a =1 a< 1 a> 1 a< 1 1 9 1 6 2 20 1 2 56 x 2 a 2 b2 x 2n 10xy n+1
(x +2y )n a x+y (a + b)x (a b)15 (a x)3n x m+1 x n+1 6a 2x +8a x 10a 2x y x 3n + y 3m a 4n + a 2n +1
31 · 5x x n+1 (x 1)2 (x 2 9y 2 )4n
(a 2 +2a +4)m (a 2)m (a 2)n (a +2)n =(a 3 8)m (a 2 4)n 7a n b 3x 2b y 2(a b) 2a(1 a)
(x + y )2n (x y + z )n m 8an xa 21by 5a2 b2(n 1) 4xy 259 216 · 6x x a x b (x a + x b + x 2 ) x a
(x 1)2 (a b)m+n 1
(a + b)(x y ) a + b x y 4axy (4a)xy 102m 3x 7y
(x m n )m+n ((x + y )a+2 )a ((x y )a 2 ab+b2 )a+b x 8 15x 8 y 5 by 2cz
2x 3 5x 2 +18 a 2x ± 2a x b x + b 2x a 3
b4 a2 = a 2 b4 2xy 6 3x 4 y x 24 a5 y 12 b7 x13 1+ x 1 x 2 n = 4 n = 10 n = 2 n =1 n =7 n =1 9 109 3, 792 1010 4, 2073 1011 6 10 2 1, 7 10 4
6, 3 · 10 6 2, 0125 · 10 7 1, 1 · 10
5, 98 109 1 3 1 2 x 17 y 50 a 12 b3 3 5
2n +2n =2 · 2n =2n+1 21 n 2 n =2 · 2 n 2 n =2 n 2 · 3n +3n =3 · 3n =3n+1 2 n +2 n+1 =2 n +2 · 2 n =3 · 2 n
3n 2 n 0, 163
1+2x f (3)= f ( 3) f (5) >f (0) f ( 8) >f (0) f ( 6) <f (8) f ( 3) >f ( 1) f (4) <f (7) h(4) >h( 4) h( 10) <h(0) h(12) >h(0) h( 82) <h(45) h( 5) <h( 2) h(7) <h(9) 78 < 98 0, 48 > 0, 38 0, 98 < 18 ( 6)8 < ( 78 ) 1 3 8 > 1 4 8 ( 1)8 < ( 1, 14)8 35 < 45 0, 65 < 0, 75 0, 725 < 1 ( 4)5 > ( 5)5 1 3 5 < 1 4 5 (0, 9) > ( 1)5 f ( a)=80 g ( b)= 121 n =3 n =2 n =4 n =5
310 < 410 1, 714 < 1, 814 0, 57 > 0, 47 1 3 17 > 1 5 17 4 9 10 = 2 3 20 321 > 87
(a 1)(a2 +4) a(1 4a)
y x x> 0 y> 0
x 2 y 3 x> 0 y> 0 x = y
6
(a 4 )2 (a 30 )2 (a 10,5 )2 (a 8 )2 (a 3,5 )2 (a 1,5 )2 (a 0,125 )2 (a 1 2 )2 (a 1 6 )2 (x 2 )3 (x 6 )3 (x 5 3 )3 (x 1 3 )3 (x 1 6 )3
)
xy 1 2 + x 1 2 y d 1 3 c 1 3 d 2 3 c =0 x x 1 3 +2x 2 3 2 a 1 +2a 1 4 3a 3 4 6 a 2 3 2a 1 3 b 1 3 + b 2 3 x 2x 1 2 y 1 2 + y x 1 2 + y 1 2 x 1 2 y 1 2 x 1 2 + y 1 2 1 x 1 2 +3 1 x 3 a 1 3 b 1 3
y + x y x (1 y )2 x 1 2 a 2 3 a 2 3 x 1 2 + y 1 2 x 1 2 y 1 2 +1
√b
x
x 3 x = y 2x 1 x =16 x> 0 a = b a2 b2 a + b = (a b)(a + b) (a b)= a b ( a + b a b) x =0 = ±27 a 2 = b3 b> 0
x x> 0 √6x x> 0 2 √a 2+ √a +2 |x|√2 |x| 1
((1+ i)(2 i)) (3+2i)=(1+ i) ((2 i)(3+2i))
(x 2i)(x +2i) (3x +4i)(3x 4i) (5x + i)(5x i) (x + √2i)(x √2i) (√3x + i)(√3x i) (i +1)4 (1 i)4 =1+4i +6i2 +4i3 + i4 (1 4i +6i2 4i3 + i4 )
=8i +8i3 =8i 8i =0 i + i2 + i3 + i4 =0 i4k +1 = i i4k +2 = 1 i4k +3 = i z = i +1 i 1 i +1 i +1 n = Å (i +1)2 i2 1 ãn = (i2 +2i +1)n ( 1 1)n = (2i)n ( 2)n =( 1)n in n
3x 2 12x +3 x 2 +2x +9 2x 2 x 6 4x 2 3x 8 3x 2 +2ix +2 5i x 2 4ix +2 3i P2 (0)=1 P2 (1)=3 P2 (i)= i x 2 a + b =1, a b =1 a =1,b =0 (x 1)(x 8) (x 5)(x +12)
(x)=
x 2 + x 2=0 x1 = 2 x2 =1 x2 =1 x 0 x 2 x 2=0 x1 =2 x2 = 1 x2 = 1 x< 0 1 1 √13 2
x1 + x2 =6 x1 x2 =8 x1 x2 = 2 3 x1 + x2 = 7 3 x 2 5x +6=0 x 2 + x 2=0 x 2 7x +12=0
x 2 (√3+ √5)x + √15=0 x 2 2x 1=0
k = 5
(x 9)(x 1) (x +6)(x 10) (x +6)(x +19)
3x +1
x +1 m n m + n = a mn =1 b a 2 + b2 =(m + n)2 +(1 mn)2 =(m 2 +1)(n 2 +1) ax 2 + bx + c =0 a b c √2 1 √2 √2+ 1 √2 = b a (a =0,b =0)
2= 3a b √2 x1 x2
x1 + x2 = 2 5 x1 x2 = 3 5 y1 y2 y1 = x1 +7 y2 = x2 +7 y1 + y2 =(x1 +7)+(x2 +7)=(x1 + x2 )+14= 2 5 +14=14 2 5 y1 y2 =(x1 +7)(x2 +7)= x1 x2 +7(x1 + x2 )+49= 3 5 +7 · 2 5 +49=51 1 5 y 2 14 2 5 y +51 1 5 =0 5y 2 72y +256=0 y = x +7 x = y 7 5(y 7)2 2(y 7) 3=0 5y 2 72y +256=0 x1 + x2 =8 x1 x2 = c x1 =3x2 x1 + x2 =8, x1 =3x2 x1 =6 x2 =2 c =6 · 2=12 x1 + x2 =5,x1 x2 =6, (x1 + x2 )2 = x 2 1 + x 2 2 +2x1 x2 x 2 1 + x 2 2 =(x1 + x2 )2 2x1 x2 =52 2 · 6=13 x1 x2 x 3 1 + x 3 2 =(x1 + x2 )(x 2 1 x1 x2 + x 2 2 )=(x1 + x2 )((x1 + x2 )2 3x1 x2 ) = p(p 2 3q )= p(3q p 2 ) x1 + x2 = p x1 x2 = q
x 3 1 + x 3 2 = p(3q p 2 )
+1)
+1)
m2 =4 m2 = 4 {1;4}; {1; 4}
x y (2;1);(2; 1);( 2;1);( 2; 1) x1 =9 y1 =4 x2 = 9 y2 =
=4 x4 =9 y4 = 4 x1 =3 y1 =4 x2 =3 y2 = 4 x3 = 3 y3 =4 x4 = 3 y4 = 4 x1 =4 y1 =3 x2 =4 y2 = 3 x3 = 4 y3 =3 x4 = 4 y4 = 3 (6;4), (6; 4), ( 4; 6);( 4; 6) (2;3), (2; 3), ( 2;3), ( 2; 3)
= b2 4ac =72 4 3 4=1 > 0 x1 + x2 = b a = 7 3 < 0 x1 x2 = c a = 4 3 > 0 D =25 x1 x2 = c a = 2 2 = 1 < 0 x1 + x2 = b a = 3 2 > 0 |x1 | > |x2 |
D =144 > 0 (D> 0) x1 x2 = c a > 0 (b< 0)
D =0
= 4 < 0
9 4 · 2 · (k 1) 0 k 1 2 > 0 3 2 > 0 1 <k 17 8 1 <k 17 8
D =36(4 k ) D> 0 k< 4 D =0 k =4 D< 0 k> 4 D =(k +1)2 4(2k 1)=0
=9 x2 =51 y2 = 34 8 9 51 34 x y xy =520, (x 3)(y +12)=520 x
=
y
y
Å b 2a ; 4ac b2 4a ã T 1 4 , 49 8 x1 =2 x2 = 3 2 T 3 4 ; 31 8 T (1, 2) x 10 x y y = x(10 x) y =10x x 2 a = 1 < 0 x = b 5 2a y = 4ac b2 4a =25 x 10 x c 2 = x 2 +(10 x)2 y = c 2 y =2x 2 20x
x y P
y = 4R2 x2 P = xy = x 4R2 x2
P 2 = x 2 (4R2 x 2 ) x 2 = u P = v
v = u 2 +4R2 u u =2R2 x = R√2 y x y
x + y 3x +2y = O (x + y ) · √3 2 x = P. O =200 3x +2y =200
P = x2 √3 2 +50x√3 x = b 2a =50 x + y =75 a b c
O = a 22 + b 2+ c
7= a 32 + b 3+ c
8= a( 2)2 + b( 2)+ c
a = 1 b = 2 c =8
f (x)= x 2 2x +8
f (x)=3x 2 +11x 9
f (x)= x 2 6x +5
q =3
p = 2 q =4
k = 3
x ∈ (−∞, 2) x ∈ (2, ∞) x ∈ (−∞, 1) x ∈ 1, ∞) x ∈ −∞, 1 4 x ∈ 1 4 , ∞ x1 = 2 x2 =2 y (0)=4 x ∈ (−∞, 0) x ∈ (0, ∞) T (0; 4) x1 = 1 x2 = 1 2
1 4 ; 9 8 x1 = 2 x2 = 3 5 y (0)=6 x ∈
T 1 4 , 49 8 f (x)= ax 2 + bx + c f (0)= c< 0 Ox f (1)= a + b + c> 0
,
Ox ax 2 + bx + c =0 |x| = x,x 0 x,x< 0
x2 =1 x ∈ (−∞, 2) x ∈ (1, ∞) S =(−∞; 2) ∪ (1, ∞)
ax 2 + bx + c = a(x x1 )(x x2 ) x 2 + x 2=(x +2)(x 1) (x +2)(x 1) > 0 (x +2 > 0 ∧ x 1 > 0) ∨ (x +2 < 0 ∧ x 1 < 0) x +2 > 0 x 1 > 0 x +2 < 0 x 1 < 0, x>
2 x> 1 x< 2 x< 1 x> 1 x< 2 x (−∞, 2) 2 ( 2, 1) 1 (1, ∞) x +2
x 1
+ (x +2)(x 1) + 0
+ x + x ∈ R 2 x 1 2 x|− 1 <x< 1 4 x ∈ R S = ∅ x S =( 5; 2) ∪ (3; ∞) x< 3 x 1 x +2 x = 2 x = 2 x +2)2 > 0 (x +2)2 (x 1)(x +2) > 0 ∧ x = 2 (x 1)(x +2) > 0 (−∞, 2) ∪ (1, ∞) x = 2 5 3x +7 2 5x > 1
3x +7 2 5x +1 > 0 2x 9 5x 2 > 0 x = 2 5 ∧ 2x 9 5x 2 > 0 −∞; 2 5 ∪ 9 2 ; ∞ x|− 1 <x< 1 2 x ∈ R x 2 2x + k 3 > 0 D< 0 4 4(k 3) < 0 (k> 4) x ∈ ∅
D> 0 D =(k 2)2 4 > 0 k (k 4) > 0 k< 0 k> 4 D = k 2 4 < 0( 2 <k< 2) x (k 1) > 0 D = k 2 (k 1)(3k 2) < 0 k 1 > 0 k 1 > 0, 2k 2 5k +2 > 0 k> 1 k< 1 2 k> 2 k> 2
x =4 x 7 ∧ x 2 2x 99=0 x =11 2x 2 +7=(x 2 4)2 , x 2 4 0 x 4 8x 2 +16=0 ∧ (x 2 ∨ x 2) x1 =3 x2 = 3 x =3 x 2 6x +7=0 x 2 4x +13=0 D =(2(k +1))2 4(k 1)(k 2)=0 k = 1 5 x1 + x2 = 1 2k 2 x1 x2 = k 1 2 3x1 4x2 =11 k1 = 2 k2 = 33 8 u 2 2u 63=0 u = x 3 16x u1 =9 u2 = 7 9= x 2 16x 7= x 2 16x 8+ √73, 8 √73, 8+ √57, 8 √57
2i = ±(1+ i) x1 =6 3i x2 =5 4i 22, 21, 20 b + c =5 a bc =8 a(b + c) b + c =5 a bc =8 a(5 a) b c x 2 (5 a)x +(a 2 5a +8)=0 1 a 7 3 b c
2 <m< 0 55 52 (2m +1)x 2 2(3m +4)x (3m +4)=0
D =(3(2m 1))2 0 x1 = 2 m m +1 x2 = m +1 2 m (m = 1, 2) m = 1 2 x1 = x2 =1 m = 17 7 x =1 x 2 = u y 2 = v u + v =25, u +2v =11 x 2 = u y 2 = v u =9 v =16 x 2 =9 y 2 =16 {(3, 4), (3, 4), ( 3, 4), ( 3, 4)} ax 2 + by 2 = c, a1 x 2 + b1 y 2 = c1
{(4, 1), (4, 1), ( 4, 1), ( 4, 1)}
x 2 + αx β =0 D = α 2 +4β
α + β =1 D =(α 2)2 p q p q
ax 2 + bx + c =0 a p q 2 + b p q + c =0
p q p q p q p q a bc
ap 2 + bpq + cq 2 =0
ap 2 + bpq + cq 2
ap 2 + bpq + cq 2 =0
3(x1 + x2 )2 x1 x2 4x1 x2 (x1 + x2 )
D =1+4k (k =2, 6, 12,...)
p = q =0 p =1 q = 2 x1 + x2 =2k +1 x1 x2 = k 2 +2 x1 = 1 2 x2 (k =4)
(x √5)(x + √5)(x 3i)(x +3i)=0 x 4 +4x 2 45=0 x 2 + x +1= u (x1 = 2 x2 =1 x3,4 = 1 ± i√19 2 (x 2 11x +28)(x 2 11x +30)=1680 x 2 11x +30= u x1 = x2 x3 = x4 x1 + x2 + x3 + x4 =0 x + y + xy =11, (x + y )xy =30. x + y = u xy = v u + v =11, uv =30 u1 =6 v1 =5 u2 =5
v2 =6 x + y =6, xy =5 x + y =5, xy =6
{(5, 1), (1, 5), (3, 2), (2, 3)} x 2 y 2 = u xy = v x(x +1)= u 3x 2 +5y = v y y =0 x =0 x y 2 3 x y +2=0 x y = u u 2 3u +2=0
u1 =2 u2 =1 x y =2 x 2 3x y +3=0 x y =1 x 2 3x y +3=0
{(2, 1), ( 3 2 , 3 4 ), (1, 1), (3, 3)}
ax 2 + bxy + cy 2 =0
(x +1) (x 2 x +1)=0 x1 = 1 x2 = 1+ i√3 2 x2 = 1 i√3 2
2 ,f3 ,f4 ,f8
3+3 x +2 (x 2) 2 y (x y ) (x + y ) 12 x 8 y 1 3 ( x +2 y ) 1 2 · ( x + 1 2 ( x + 1 2 ( x + 1 2 x))) (x 1) x 1 m (n x + q p y ) x 2 x x x x
)= a c ⇔ x = a c 1+ a b ⇔ x = 1 1 a c + a b a c = 1 1 a c + 1 b c b c = a c
f 1 = {(2, 1), (3, 2), (3, 5), (4, 4)}
f f 1 = {(0, 0), (1, 1), (2, 2)}
f 1 (x)=2x f 1 (x)= x 1
f 1 (x)= 1 3 x + 17 3 f 1 (x)= 1 7 x 5 7 x 0 (0, +∞) (1, +∞)
f 1 (x)= √x 1
x 0 f (−∞, 0) −∞, 0) f 1 (x)= √x ( 1, 1) f ∩ f 1 = {(x,x)|x ∈ R}
f ∩ f 1 = {(x, x)|x ∈ R} f ∩ f 1 = ∅
f 1 (x)= 1 2 · x + 3 2 f 1 (f (x))= f 1 (2x 3)= 1 2 · (2x 3)+ 3 2 = x
f 1 (x)= x f 1 (f (x))= f 1 (x)= x
f 1 (x)= x 2 x 0 f 1 (f (x))= f 1 (√x)=(√x)2 = x
f 1 = {(1, 1), (3, 2), (4, 3)} f 1 (f (1))= f 1 (1)=1
f 1 (f (2))= f 1 (3)=2 f 1 (f (3))= f 1 (4)=3
Df = x ∈ R|x> 3 5 f (x)=0 x = 4 5
Df = x ∈ R| 5 3 <x< 2
Df = {x ∈ R|1 <x< 2} f (x)=0 x = 3 2
f = R f (x)=0 x =0
D
f 1 (x)=3 x 2 4 f 1 (x)= 1 2 · 3x + 1 2
f (x) > 0 x ∈ 5 2 , 2 f (x) < 0 x ∈ ( 2, +∞)
f (x) > 0 x< 0 f (x) < 0 0 <x< 1
f (x)= 2 x, x> 0 2 ( x), x< 0
6
4= 10 2+ 10 2=0, 60206
2, 5= 10 10 4 = 10 10 10 4=1 0, 60206=0, 39794 10 5=0, 69897 10 8=3 10 2 7 3= 10 3 10 7 ≈ 0, 564575 x ≈ 1, 9054607 x ≈ 0, 037634
,
R =15, 669852 n =11 n< 3700 2 <n +1 n b a <n +1 n b a
684, 366Ω s v =11, 8466 t =6, 04 x =1 x
4a2 √3 9 2 b = H 2 + Å a√3 6 ã2 h = H 2 + Å a√3 3 ã2 a 2 a√3 2 H =12 H =12 aH a + H =3 3 7 a 4 x a 2 : x 2 = Ä a 4 ä2 : Ä a 4 xä2 x = a 5 a 4 3b2 a2
P = a√3a2 +144H 2 48
P = a√a2 +8H 2 8 2 2 3√579 4 2 H = a√6 3 a = H √6 2 a = 2P (6 √3) 33 a = P (2 √3) 3
a2 √3 2 +3ab √2 2
P =768 2
H = a 2
M = a2 √15 4
M =3√39 2
M =540 2
M =396 2
V = d3 √3 9
V =6 3
a =30
V =4500 3 2
H =4 3
V =100 3
V =121, 5 3
V =128 3
V =2, 5 3 V =10 3
a =3… 2V 3 β =60◦ B B1 = H 2 H 2 1
V V1 = 3BH 3B1 H1 = BH B1 H1 = B B
V = Q 2 Q√2 a =28 b =36 c
P =478 2
P =700 2
a3 2 (2 α) 2a(a + 4b2 a2 )
V = ad√12a2 3d2 8 2 2
V = 1 2 2BQQ1 ≈ 169 2
V =192 3
P = a2 (√3+2) 2 2
B = b2 4 √3 2 M = b2 (4√3+ √15) 4 2
x = H 3 √1+ m
1:7:19:37:61 2
P =764 2 V =1389 3
P =114, 32 2 V =78, 7 3
3:4
r =3, 5 l =12, 5 1:9 4:9 x =4 a =2√2
1:2
H = r 2
rπs =2r 2 π =2B B + M =3B
P = πH 2
s =2H
M =2πQ
M =2π (R2 r 2 )
= r 2 π M =
Mk = rπs =2rπH Mv =2rπH
60◦ M1 =2rπH
2 = rπ r 2 + H 2 2H : r 2 + H 2
V =4π √2 3
V2 <V3 <V1
r =3 H =3√3
V =243π 3
V =96π 3
V =12π 3
3 … 3 2 :1
4:1
V = 3πa3 4
P =208, 8 2
H =2 6 √3888
V1 : V2 =3 4 √3:4
V = 2π 9 3
3:4
V1 : V2 =8:3
V = πh2 c 3
x = r 2 H R2 + r 2
r1 = 3 … R3 + r 3 2
V1 = 7 12 v
r =6 P =16π 2 V = 32 3 π 3 P =48π 2 V =32√3π 3
P =196π 2 V = 1372π 3 3 r H =2r
M =2rπ · 2r =4r 2 π
P =910π 2 V1 = 11968π 3 3 V2 = 37532π 3 3
R =3
P = 2πR2 H H + R = 2π 63702 60 6430 ≈ 2379000 2
2r 1 3
V = 11πR3 24 R = a√21 26 R = a√219 24 H = 24 13 R a a a 2a 2 π +2aπa =4a 2 π a
V = 4000π 3 3
P = 400π 9 2 V = 4000π 81 3
≈ 4, 6 r =9 H =18 1458π 3
V = a3 4π
V1 : V2 = a : b
V = a3 4
r 2 H Åπ 3 √3 4 ã
4r 2 2√3 3
2π :9√3
M =2πa 2 =200π 2
V1 : V2 : V3 =41:58:50
P = a 2 π √3 V = a3 π 4
H = M
2π (P M )
x = r (2H r ) H
P = p2 4π (1+ √2) V = p3 24π 2
25:36
P1 : P2 =4:5 V1 : V2 =6:5√5 R = 1 2 a2 +
P =70π
60◦ P = 7a2 π 2 V = 7a3 π √3 12
P =2a 2 π √3 V = a 3 π
M =1666π 2 V = √3 3 74431 3
P = √2 2 πc 2 2
V = πa 3 α 2α
s =3 r = 3(√5 1) 2 H
V = 1900π 27
P =8a 2 π √2 V =2a 3 π √2 a P =3a 2 π a =6
V = a3 π √3 4 =54π √3
V2 : V1 =63
V =96π 3
r = Ha√3 6 …h2 + a2 12
πH (R r )2 12
(3π /4)= (π /4)= √2/2 (3π /4)= (π /4)= √2/2
(3π /4)= (π /4)= 1 (3π /4)= (π /4)= 1 (5π /6)= (π /6)=1/2 (5π /6)= (π /6)= √3/2
(5π /6)= (π /6)= √3/3 (5π /6)= (π /6)= √3
(5π /4)= (π /4)= √2/2 (5π /4)= (π /4)= √2/2
(5π /4)= (π /4)=1 (5π /4)= (π /4)=1
(7π /6)= (π /6)= 1/2 (7π /6)= (π /6)= √3/2
(7π /6)= (π /6)= √3/3 (7π /6)= (π /6)= √3
(5π /3)= (π /3)= √3/2 (5π /3)= (π /3)=1/2
(5π /3)= (π /3)= √3 (5π /3)= (π /3)= √3/3
(7π /4)= (π /4)= √2/2 (7π /4)= (π /4)= √2/2
(7π /4)= (π /4)= 1 (7π /4)= (π /4)= 1
390◦ = 30◦ =1/2 405◦ = 45◦ = √2/2
420◦ = 60◦ = √3 540◦ = 180◦
450◦ = 90◦ =1 390◦ 30◦ = √3/2
630◦ = 270◦ 690◦ = 330◦ = 30◦ = √3
(9π /4)= (π /4)= √2/2 (17π /6)= (5π /6)= (π /6)=1/2
(11π /3)= (5π /3)= (π /3)= √3 (19π /2)= (π /2)=0
(13π /3)= (π /3)= √3/2 (15π /4)= (7π /4)= (π /4)= 1
(10π /3)= (4π /3)= (π /3)= √3/3 (13π /6)= (π /6)= √3/2
( 30◦ )= 30◦ = 1/2
( 60◦ )= 60◦ =1/2
( 45◦ )= 45◦ = 1 ( 90◦ )= 90◦ =0
( 720◦ )= 720◦ = 0◦ =0
( 1140◦ )= 1140◦ = 60◦ =1/2 ( 900◦ )= 900◦ = 180◦ =0
( 750◦ )= 750◦ = 30◦ = √3/3
30◦ 60◦ 90◦ =1/2 1/2 0=0
(90◦ 45◦ )+ (45◦ +45◦ ) (180◦
(π /3+ π /6)= (π /2)=1
(π /3)+ (π /6)= √3/2+1/2=(√3+1)/2 > 1 (π /3+ π /6)= (π /2)=0
(π /3)+ (π /6)=1/2+ √3/2=(√3+1)/2 > 0
60◦ + 45◦ =1/2+ √2/2=(√2+1)/2 > 1
60◦ + 60◦ = √3/2+1/2=(√3+1)/2 > 1
60◦ /( (15◦ +30◦ ) 30◦ )= 60◦ /( 45◦ 30◦ ) =(√3/2)/(√2/2 1/2)= √3/(√2 1)= √6+ √3
2 ( 270◦ ) (1/2) 180◦ ( 90◦ )=2 270◦ (1/2) 180◦ + 90◦ =0 0+1=1
( 45◦ )+ ( 45◦ )= 45◦ + 45◦ = √2/2+ √2/2=0 ( 90◦ )+ ( 90◦ )= 90◦ + 90◦ = 1+0= 1 ( 360◦ )+ ( 360◦ )= 360◦ + 360◦ =0+1=1 ( 180◦ )+ ( 180◦ )= 180◦ + 180◦ =0 1= 1 ( 420◦ )+ ( 420◦ )= 420◦ + 420◦ = 60◦ + 60◦ = √3/2+1/2=( √3+1)/2 ( 1710◦ )+ ( 1710◦ )= 1710◦ + 1710◦ =
(60◦ α)= (90◦ (60◦ α))= (30◦ +
◦
) (80◦ α/2)= (90◦ (80◦ α/2))= (10◦ + α/2) (30◦ 2α)= (90◦ (30◦ 2α))= (60◦ +2α) (α +60◦ )= (90◦ (α +60◦ ))= (30◦ α) 0, 7 α β α + β =180◦ β = (180◦ α)= α = 0, 7 k 1/k α1 β
+ β )= (π γ )= γ 2(
(α + β )= (π γ )= γ 2(α + β )= 2(π γ )= (2π 2γ )= 2γ 1 1+2 α 3 1 α 1 0 α 1 2 3 α 4 1 α 1 α 0 2 2 α 2
(γ + δ )= 84/85 (γ δ )= 36/85
(γ + δ )=13/85 (γ δ )=77/85 α = 1 2 α = 40/41 β = 1 2 β =9/41 (α + β )= 9/41 · 9/41+( 40/41) · ( 40/41)=1
(21◦ +9◦ )= 30◦ =1/2
(63◦ 18◦ )= 45◦ = √2/2
(278◦ 68◦ )= 210◦ = 30◦ = 1/2
(32◦ +58◦ )= 90◦ =0
3α 4α ((α + π /6)+(α π /6))= 2α ((π /4+ α)+(π /4 α))= (π /2)=0
) (α + β )=( α + β )/(1 α · β )=(1/2+1/3)/(1 1/2 · 1/3)=1
/4 (α + β )/2) (π /4 (α β )/2)
2 ((α + β )/2) · ((α β )/2)+2 ((α + β )/2) ((α + β )/2) =2 ((α + β )/2) · ( )(
)/2=(
x = kπ x = π /4+ kπ x = π /4+ kπ
x = π /3+ kπ x = π /6+ kπ x = π /3+ kπ x = π /6+ kπ
x = π /4+ kπ x = π /2+ kπ x = π /4+ kπ x = π /3+ kπ
x = π /6+ kπ x = π /6+ kπ x = π /3+ kπ
2x = kπ x = kπ /2
3x/2=3π /2+2kπ x = π +4kπ /3=(4kπ +3)π /3
x π /6= π /2+2kπ x =2π /3+2kπ
x +30◦ =225◦ + k 360◦ x 30◦ = 45◦ + k 360◦ x =195◦ + k 360◦
x = 15◦ + k · 360◦
x =(75◦ + k · 360◦ )/2 x =(195◦ + k · 360◦ )/2
x = π /12+2kπ x = 5π /12+2kπ
x = 11π /18+2kπ /3=(12k 11)π /18 x = 5π /18+2kπ /3=(12k 5)π /18
3x =2kπ x =2kπ /3 2x/3= π /2+ kπ x =(6k +3)π /4
x + π /3= π +2kπ x =(6k +2)π /3
60◦ 2x =60◦ + k1 · 360◦ 60◦ 2x = 60◦ + k1 · 360◦ x = k 180◦ x =120◦ + k 180◦ x/2+ π /4=5π /6+2kπ x/2+ π /4=7π /6+2kπ
x =(24k +13)π /6 x =(24k +17)π /6
2π /3 x =2π /3+2kπ 2π /3 x =4π /3+2kπ x =2kπ x =2π /3+2kπ x/3 45◦ =45◦ + k1 · 360◦ x/3
◦ = 45◦ + k1 · 360◦ x =270◦ + k 360◦ x = k 360◦
2x = π /4+ kπ x =(4k 1)π /8 x/3= π /4+ kπ x =(12k +3)π /4
π /2 x = kπ x = π /2+ kπ x/2 30◦ =60◦ + k
3π /4+ x = π /4+ kπ x =(k 1)π 120◦ +2x = 30◦ + k · 180◦ x = 75◦ + k · 90◦
3x/2 π /3= π /3+ kπ x =2kπ /3
x/2= π /4+ kπ x =(4k +1)π /8 4x = π /2+ kπ x =(2k +1)π /8
2x/3= π /4+ kπ x =(12k 3)π /8
2x π /3= π /4+ kπ x =(12k +7)π /24 x +120◦ = 60◦ + k · 180◦ x =(k 1) · 180◦
5π /6 x/2= π /6+ kπ x =(6k +4)π /3 135◦ +3x = 45◦ + k 180◦ x =(k 1) 60◦ x = √2/2 x = √2/2 x = π /4+2kπ x =5π /4+2kπ x = π /4+2kπ x =3π /4+2kπ x =(2k +1)π /4
x =(3k +1)π /3 x =(3k +2)π /3 x =(3k +1)π /3 x =(3k +2)π /3
x =(2k 1)π /4
x =(6k 1)π /6 x =(6k +1)π /6 x =(6k 1)π /6 x =(6k +1)π /6
F1 F1 =2018, 8 F1 =1350, 6
ACD
= a ACD =90◦ α F1 = Q/ 48◦ =2018, 8 F2 = Q 42◦ F1 =500 F2 =866 F1 = Q α F2 = Q α
