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Sagot :

1.
(-2x + 3)(x²- 3x + 2) ≤ 0

-2x + 3 ≤ 0
-2x ≤ -3
x ≥ 3/2


x² - 3x + 2 = 0

Δ = 9 - 4*1*2 = 9-8 = 1
x1= (3-1)/2 = 1
x2 = (3+1)/2 = 2

x ∈ [1; 3/2] U [2; ∞)

2.

(-2x² + 3x - 4)(5x² + 3x - 8) > 0

-2x² + 3x - 4=0

a= -2 alors l’inéquation est négative n’importe pour quelle valeur du x

5x² + 3x - 8 = 0
Δ = 9 - 4*5*(-8) = 9 + 160 = 169
x1= (-3 - 13)/10 = -1,6
x2 = (-3 + 13)/10 = 1

x ∈ (-1,6 ; 1)






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