Câu hỏi của Hoàng Thùy Linh

Question

tìm x biết

a)$\left(x-2\right)^2-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)=0$

b)$\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)-x^6+2x=1$

c)\(\left(x-3\ri...

Nguyễn Lê Phước Thịnh · Accepted Answer

b) Ta có: $\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)-x^6+2x=1$

$\Leftrightarrow\left(x^3-8\right)\left(x^3+8\right)-x^6+2x-1=0$

$\Leftrightarrow x^6-64-x^6+2x-1=0$

$\Leftrightarrow2x-65=0$

$\Leftrightarrow2x=65$

hay $x=\frac{65}{2}$

Vậy: $x=\frac{65}{2}$

c) Ta có: $\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1$

$\Leftrightarrow x^3-27-x\left(x+2\right)\left(x-2\right)-1=0$

$\Leftrightarrow x^3-27-x\left(x^2-4\right)-1=0$

$\Leftrightarrow x^3-27-x^3+4x-1=0$

$\Leftrightarrow4x-28=0$

$\Leftrightarrow4x=28$

hay x=7

Vậy: x=7Câu trả lời: b) Ta có: $\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)-x^6+2x=1$