Toán 8

C

chonhoi110

a) $(x+1)^2(x+2)+(x−1)^2(x−2)=12$

$\leftrightarrow (x^2 + 2x + 1)(x + 2) + (x^2 - 2x + 1)(x - 2) - 12 = 0 $

$\leftrightarrow x^3 + 2x^2 + 2x^2 + 4x + x + 2 + x^3 - 2x^2 - 2x^2 + 4x + x - 2 - 12 = 0$

$\leftrightarrow 2x^3 + 10x - 12 = 0 $

$\leftrightarrow 2(x - 1)(x^2 + x + 6) = 0 $

$\leftrightarrow x=1$ (vì $x^2 + x + 6 > 0$)
 
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