Question

giải phương trình

\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\)

Answer 1

\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\)

<=> \(\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{1}{x+1}=0\)

<=> \(\frac{2}{\left(x-1\right)^2\left(x+1\right)}+\frac{3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}+\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}=0\)

<=> \(2+3x-3+x^2-2x+1=0\)

<=> x² + x = 0

<=> x(x + 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy S = {0; -1}