bạn ghi rõ đề đi

<=>16=x²+8x+16

<=>-x²-8x=0

=>-x(x+8)=0

Th1:-x=0

=>x=0

Th2:x+8=0

=>x=-8

\(x+\dfrac{1}{x}=a;x^2+\dfrac{1}{x^2}=a^2-2\)

\(8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)a^2=\left(x+4\right)^2\)

\(8a^2+4\left(a^2-2\right)\left[\left(a^2-2\right)-a^2\right]=\left(x+4\right)^2\)

\(8a^2-8\left(a^2-2\right)=\left(x+4\right)^2\)

\(16=\left(x-4\right)^2;\left[{}\begin{matrix}x-4=4;x=8\\x-4=-4;x=0\end{matrix}\right.\)

`2/(4-x^2)+1/(x^2-2x)=(x-4)/(x^2+2x)(x ne 0,+-2)`

`<=>(2x)/(4x-x^3)+(x+2)/(x^3-4x)=(x^2-6x+8)/(x^3-4x)`

`<=>-2x+x+2=x^2-6x+8`

`<=>x^2-7x+10=0`

`<=>x^2-2x-5x+10=0`

`<=>x(x-2)-5(x-2)=0`

`<=>(x-2)(x-5)=0`

Vì `x ne 2=>x-2 ne 0`

`=>x-5=0`

`=>x=5`

Vậy `S={5}`

b) ĐKXĐ: \(x\ne1\)

Ta có: \(\dfrac{2}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{x}{x^2+x+1}\)

\(\Leftrightarrow\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

Suy ra: \(2x^2+2x+1-3x^2-x^2+x=0\)

\(\Leftrightarrow-2x^2+x+1=0\)

\(\Leftrightarrow-2x^2+2x-x+1=0\)

\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

Bn ghi đề khó nhìn quá!