a) đkxđ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\x^2-9\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
\(M=\frac{2x}{x+3}+\frac{x+1}{x-3}+\frac{11x+3}{x^2-9}\)
\(=\frac{2x\left(x-3\right)+\left(x+2\right)\left(x+3\right)+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x^2+9x}{\left(x-3\right)\left(x-3\right)}\)
\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)
b)\(M=\left|\frac{3}{2}\right|\Rightarrow\frac{3x}{x-3}=\left|\frac{3}{2}\right|\)
\(\Rightarrow\) \(\frac{3x}{x-3}=\frac{3}{2}\)(I) hoặc \(\frac{3x}{x-3}=-\frac{3}{2}\)(II)
(I)\(\Rightarrow6x=3x-9\)
\(\Rightarrow3x=-9\)
\(\Rightarrow x=-3\)(loại)
\(\left(II\right)\Leftrightarrow6x=-3x+9\)
\(\Rightarrow9x=9\)
\(\Rightarrow x=1\)
Vậy...
a, \(M=\frac{2x}{x+3}+\frac{x+1}{x-3}+\frac{11x-3}{x^2-9}ĐKXĐ:x\ne\pm3\)
\(\frac{2x}{x+3}+\frac{x+1}{x-3}+\frac{11x-3}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)+11x-3}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{3x}{x-3}\)
