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)\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

Answer 1

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=7