Question

\(\frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\).

Answer 1

\(ĐK:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)

+ pt \(\Leftrightarrow\frac{x+4}{\left(x-1\right)\left(x-2\right)}-\frac{x+1+2x+5}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)-3\left(x+2\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\left(x^2+x-12\right)-3\left(x^2-4\right)=0\)

\(\Leftrightarrow-2x^2+x=0\) \(\Leftrightarrow-x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\) ( TM )