Question

  Giải Phương Trình-7x^2+4/x^3+1=5/x^2-x+1-1/x+1

Answer 1

`(-7x^2+4)/(x^3+1)=5/(x^2-x+1)-1/(x+1)(x ne -1)`

`<=>-7x^2+4=5(x+1)-x^2+x-1`

`<=>-7x^2+4=5x+5-x^2+x-1`

`<=>6x^2+6x=0`

`<=>6x(x+1)=0`

Vì `x ne -1=>x+1 ne 0`

`=>x=0`

Vậy `S={0}`

Answer 2

ĐKXĐ: \(x\ne-1\)

Ta có: \(\dfrac{-7x^2+4}{x^3+1}=\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}\)

\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{-7x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

Suy ra: \(5x+5-x^2+x-1=-7x^2+4\)

\(\Leftrightarrow-x^2+6x+4+7x^2-4=0\)

\(\Leftrightarrow6x^2+6x=0\)

\(\Leftrightarrow6x\left(x+1\right)=0\)

mà 6>0

nên x(x+1)=0

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

Vậy: S={0}