a) ĐKXĐ: \(4x^3-9x\ne0\)
\(\Leftrightarrow x\left(4x^2-9\right)\ne0\)
\(\Leftrightarrow x\left(2x-3\right)\left(2x+3\right)\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\2x\ne3\\2x\ne-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\\x\ne-\frac{3}{2}\end{matrix}\right.\)
*Rút gọn
Ta có: \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}\)
\(=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(4x^2-9\right)}=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(2x+3\right)\left(2x-3\right)}=\frac{x\left(2x+1\right)}{2x-3}=\frac{2x^2+x}{2x-3}\)
Vậy: \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}=\frac{2x^2+x}{2x-3}\)
b) Khi M=0 thì
\(\frac{2x^2+x}{2x-3}=0\)
\(\Leftrightarrow2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\left[{}\begin{matrix}x=0\left(loại\right)\\2x+1=0\end{matrix}\right.\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)(tm)
Vậy: Khi M=0 thì \(x=\frac{-1}{2}\)
