Câu hỏi của hung

Question

tim x:

6(x+1)^2-2(x+1)^3-2(x-1)(x^2+x+1)=1

Đức Phạm · Accepted Answer

$6\left(x+1\right)^2-2\left(x+1\right)^3-2\left(x-1\right)\left(x^2+x+1\right)=1$

$\Leftrightarrow6\left(x^2+2x+1\right)-2\left(x^3+3x^2+3x+1\right)-2\left(x^3+x^2+x-x^2-x-1\right)=1$

$\Leftrightarrow6x^2+12x+6-2x^3-6x^2-6x-2-2x^3-2x^2-2x+2x^2+2x+2=1$

$\Leftrightarrow-4x^3+6x+5=0$

$\Leftrightarrow x=1.5233401602$

Câu trả lời:

$6\left(x+1\right)^2-2\left(x+1\right)^3-2\left(x-1\right)\left(x^2+x+1\right)=1$

$\Leftrightarrow6\left(x^2+2x+1\right)-2\left(x^3+3x^2+3x+1\right)-2\left(x^3+x^2+x-x^2-x-1\right)=1$

$\Leftrightarrow6x^2+12x+6-2x^3-6x^2-6x-2-2x^3-2x^2-2x+2x^2+2x+2=1$

$\Leftrightarrow-4x^3+6x+5=0$

$\Leftrightarrow x=1.5233401602$