Question

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

Answer 1

ĐKXĐ: \(\left[{}\begin{matrix}x\ne-3\\x\ne1\end{matrix}\right.\).

\(PT\Leftrightarrow\dfrac{4}{\left(x+3\right)\left(x-1\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}\)

\(\Rightarrow\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)=4\)

\(\Leftrightarrow2x^2-7x+5-2x^2-6x=4\)

\(\Leftrightarrow13x=1\Leftrightarrow x=\dfrac{1}{13}\). (TMĐK)

Vậy..