Question

Giải pt sau:

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

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\2-x\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

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

\(\Leftrightarrow x-2-5x-10=2x-3\)

\(\Leftrightarrow x-5x-2x=10+2-3\)

\(\Leftrightarrow-6x=9\)

\(\Leftrightarrow x=\dfrac{-3}{2}\) ( thỏa mãn)

Vậy nghiệm của pt là \(x=\dfrac{-3}{2}\)