a) ĐKXĐ: x∉{-1;1}
Đặt \(\frac{3x^2+x-2}{x^2-1}=0\)
\(\Leftrightarrow3x^2+x-2=0\)
\(\Leftrightarrow3x^2+3x-2x-2=0\)
\(\Leftrightarrow3x\left(x+1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=\frac{2}{3}\left(tm\right)\end{matrix}\right.\)
Vậy: \(x=\frac{2}{3}\) thì \(\frac{3x^2+x-2}{x^2-1}=0\)
b) ĐKXĐ: \(x\ne-\frac{3}{2}\)
Đặt \(\frac{x^2+2x^2+x+2}{2x+3}=0\)
\(\Leftrightarrow x^2+2x^2+x+2=0\)
\(\Leftrightarrow x^2\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2+1\ge1>0\forall x\)(2)
Từ (1) và (2) suy ra x+2=0
hay x=-2(tm)
Vậy: x=-2 thì \(\frac{x^2+2x^2+x+2}{2x+3}=0\)
c) ĐKXĐ: x≠1
Đặt \(\frac{x^2-1}{x^3-1}=0\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy: Khi x=-1 thì \(\frac{x^2-1}{x^3-1}=0\)
