Question

Giaỉ phương trình sau ;

2/x^3-x^2-x+1 = 3/1-x^2 - 1/x+1 

ghi cả kết luận ra nx nha ^^

Answer 1

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

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

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

\(\Leftrightarrow-\left(x^2-x+2x-2\right)=2\)

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

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

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

Vậy: S={0}