\(a,\dfrac{x+1}{x-1}+\dfrac{2x-1}{x}=2-\dfrac{x}{x\left(x-1\right)}\) (1)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\x\ne0\end{matrix}\right.\)
\(pt\left(1\right)\Leftrightarrow\dfrac{x\left(x+1\right)}{x\left(x-1\right)}+\dfrac{\left(2x-1\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{x\left(x-1\right)-x}{x\left(x-1\right)}\)
\(\Rightarrow x^2+x+2x^2-3x+1=x^2-x-x\)
\(\Leftrightarrow x^2+x+2x^2-3x-x^2+x+x=-1\)
\(\Leftrightarrow2x^2=-1\) ( vô lý )
⇒ PT vô nghiệm
\(b,1+\dfrac{1}{x+2}=\dfrac{12}{x^3+8}\) (2)
\(ĐKXĐ:x\ne-2\)
Pt(2) \(\Leftrightarrow\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^2+8}=\dfrac{12}{x^3+8}\)
\(\Rightarrow x^3+8+x^2-2x+4-12=0\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(x^2-2x+x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\left(t/m\right)\)
Vậy .............................