a) \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\dfrac{x\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x-7x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow x^3-4x^2+3x-x^3-3x^2=3x-7x^2\)
\(\Leftrightarrow-7x^2+3x-3x+7x^2=0\)
\(\Leftrightarrow0x=0\)( luôn đúng )
Vậy \(x\in R\left(x\ne\pm3\right)\)
b) \(\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}-\dfrac{1}{x+1}\)ĐKXĐ : \(x\ne-1\)
\(\Leftrightarrow\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\Rightarrow2x-1=2x+2-x^2+x-1\)
\(\Leftrightarrow2x+2-x^2+x-1-2x+1=0\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow x^2-2x+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{t/m ĐKXĐ}\right)\\x=-1\left(\text{không t/m ĐKXĐ}\right)\end{matrix}\right.\)
Vậy....
