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a fost răspuns

(x-1)²+7≥(x+4)²
(x+3)(x-2)≥(x+2)(x-3)

Răspuns :

Mikkka
x² -2x +1 +7 ≥ x² +8x+16
x²-2x-8x-x²≥16-8
-10x≥ 8
10x≤ -8
x≤ -8/10
x≤ -0,8 ⇒ x∈(-∞ ; -0,8]

x²-2x+3x-6 ≥ x²-3x+2x-6
x²+x-6≥x²-x-6
x²-x²+x+x≥-6+6
2x ≥ 0
x≥0 ⇒ x∈[0;+∞)


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