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Ch11 25
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Ch11 25
1.
Exercice 25
1 x+1 x−2 1 x+1 3 −2 0 3 −2 x 2 2 x 2 = −4 + 0 + 6(x − 2) + 2x(x − 2) − 0 − 6(x + 1) ≥ x − 19 = −4 + 6x − 12 + 2x 2 − 4x − 6x − 6 ≥ x − 19 = 2x 2 − 5x − 3 ≥ 0 x=3 Comme 2x 2 − 5x − 3 = 0 x = −1 2 On a 2(x − 3) x + 1 2 ≥0 -¥ + -1/2 - 3 + +¥ Donc x ∈] − ∞; −1 ∪ [3; +∞ [ 2
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