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02 emgega material de estudio semana 2

Page 1

7π‘₯ + 9 = βˆ’12 π‘₯ = βˆ’3 π‘₯2 + 4 π‘₯ + 3 = 0 π‘₯1 = βˆ’3

π‘₯2 = βˆ’1

7π‘₯ + 9 ≀ βˆ’12 π‘₯ ≀ βˆ’3

π‘Ž, 𝑏

π‘₯

(𝒂, 𝒃) = {𝒙 | 𝒂 < 𝒙 < 𝒃}

(π‘Ž, ∞) = {π‘₯ | π‘Ž < π‘₯}

[𝒂, 𝒃] = {𝒙 | 𝒂 ≀ 𝒙 ≀ 𝒃}

(βˆ’βˆž, 𝑏) = {π‘₯ | π‘₯ < 𝑏}

(𝒂, 𝒃] = {𝒙 | 𝒂 < 𝒙 ≀ 𝒃}

[π‘Ž, ∞) = {π‘₯ | π‘Ž ≀ π‘₯}

[𝒂, 𝒃) = {𝒙 | 𝒂 ≀ 𝒙 < 𝒃}

(βˆ’βˆž, 𝑏] = {π‘₯ | π‘₯ ≀ 𝑏}


7π‘₯ + 9 ≀ βˆ’12 (βˆ’βˆž, 3]

π‘Žβ‰€π‘ π‘Žβ‰€π‘

𝑐>0

π‘Žπ‘ ≀ 𝑏𝑐

𝑐

π‘Žβ‰€π‘ 𝑐<0 π‘Ž+𝑐 ≀ 𝑏+𝑐

7π‘₯ + 9 ≀ βˆ’12

7π‘₯ ≀ βˆ’21

π‘₯β‰€βˆ’

21 = βˆ’3 7

(βˆ’βˆž, 3]

π‘Žπ‘ β‰₯ 𝑏𝑐


π‘₯ 2 + 2π‘₯ βˆ’ 3 ≀ 0 π‘₯ = βˆ’3 π‘₯ = 1 π‘₯ 2 + 2π‘₯ βˆ’ 3 = (π‘₯ + 3)(π‘₯ βˆ’ 1) ≀ 0

(π‘₯ + 3)

(π‘₯ βˆ’ 1)

(π‘₯ + 3)(π‘₯ βˆ’ 1) ≀ 0

(π‘₯ + 3) ≀ 0

(π‘₯ βˆ’ 1) β‰₯ 0

(π‘₯ + 3) β‰₯ 0

(π‘₯ βˆ’ 1) ≀ 0.

π‘₯ ≀ βˆ’3

π‘₯β‰₯1

π‘₯ β‰₯ βˆ’3

π‘₯ ≀ 1. π‘₯ ∈ (βˆ’βˆž, βˆ’3] ∩ [1, ∞) π‘₯ ∈ [βˆ’3, ∞) ∩ (βˆ’βˆž, 1] π‘₯ ∈ βˆ… βˆͺ [βˆ’3,1]

π‘₯ ∈ [βˆ’3,1]


ЁЭБТ = ЁЭБС 2 + 2ЁЭБС Ρ‚Π˜Π’ 3 ЁЭБС1 = Ρ‚Π˜Π’3 [Ρ‚Π˜Π’3,1]

ЁЭБС2 = 1


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02 emgega material de estudio semana 2 by PDLM - Issuu