Toán 9 tìm MIn

V

vipboycodon

$A = (x^2-x)(x^2+3x+2)$
= $x(x-1)(x+1)(x+2)$
= $(x^2+x)(x^2+x-2)$
Đặt $t = x^2+x$
ta có: $A = t(t-2) = t^2-2t = t^2-2t+1-1 = (t-1)^2-1 \ge -1$
Vậy Min A = $-1$ khi $t = 1 \leftrightarrow x^2+x = 1 \leftrightarrow x = \dfrac{-1 \pm \sqrt{5}}{2}$
 
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