Question

giải pt chứa ẩn ở mẫu

a) \(\frac{x}{x+1}\)-\(\frac{2x-3}{x-1}\)=\(\frac{2x+3}{x^{2-1}}\)

b) \(\frac{x-1}{x}\)-\(\frac{x-2}{x+1}\)=2

Answer 1

a) \(\frac{x}{x+1}-\frac{2x-3}{x-1}=\frac{2x+3}{x^2-1}\) \(\left(ĐKXĐ:x\ne\pm1\right)\)

\(\Leftrightarrow x\left(x-1\right)-\left(2x-3\right)\left(x+1\right)=2x+3\)

\(\Leftrightarrow x^2-x-2x^2-2x+3x+3=2x+3\)

\(\Leftrightarrow-x^2-2x=0\Leftrightarrow-x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Answer 2

b) \(\frac{x-1}{x}-\frac{x-2}{x+1}=2\) \(\left(ĐKXĐ:x\ne0;x\ne-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-x\left(x-2\right)=2x\left(x+1\right)\)

\(\Leftrightarrow x^2-1-x^2+2x=2x^2+2x\)

\(\Leftrightarrow2x^2=-1\left(\text{vô lí}\right)\)

Vậy phương trình vô nghiệm.