Так как
[m]x^3-7x^2-7x+1=(x^3+1)-(7x^2+7x)=(x+1)(x^1-x+1)-7x(x+1)=[/m]
[m]=(x+1)(x^2-x+1-7x)=(x+1)(x^2-8x+1)[/m]
[m] x^3-8x^2+x=x\cdot (x^2-8x+1)[/m]
Тогда
[m]\frac{6}{(x+1)(x^2-8x+1)}-\frac{8}{x(x^2-8x+1}=\frac{1}{x^2+x}[/m]
[m]\frac{6x-8\cdot (x+1)}{x(x+1)(x^2-8x+1)}=\frac{1}{x\cdot (x+1)}[/m]
[m]\frac{6x-8x-8}{x(x+1)(x^2-8x+1)}=\frac{1}{x\cdot (x+1)}[/m]
[m]\frac{-2x-8}{x(x+1)(x^2-8x+1)}=\frac{1}{x\cdot (x+1)}[/m]
x ≠ 0
x+1 ≠ 0
[m]-2\cdot (x+4)\cdot x\cdot (x+1)=x\cdot (x+1)\cdot (x^2-8x+1)[/m]
[m]x\cdot (x+1)\cdot (x^2-8x+1)+2\cdot (x+4)\cdot x\cdot (x+1)=0[/m]
[m]x\cdot (x+1)\cdot (x^2-8x+1+2x+8)=0[/m]
[m]x\cdot (x+1)\cdot (x^2-6x+9)=0[/m]
x ≠ 0
x+1 ≠ 0
x^2-6x+9=0
x=3
О т в е т. [b]3[/b]