Câu hỏi của __HeNry__

Question

tìm x:

a) $\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=5$

b) $\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11$

Khôi Bùi · Accepted Answer

a ) $\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=5$

$\Leftrightarrow x^3-27-x\left(x^2-16\right)=5$

$\Leftrightarrow x^3-27-x^3+16x=5$

$\Leftrightarrow16x-27=5$

$\Leftrightarrow16x=32$

$\Leftrightarrow x=2$

Vậy $x=2$

b ) $\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11$

$\Leftrightarrow x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11$

$\Leftrightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11$

$\Leftrightarrow12x-133=11$

$\Leftrightarrow12x=144$

$\Leftrightarrow x=12$

Vậy $x=12$Câu trả lời: a ) $\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=5$