Question

giải p.trình sau: \(\dfrac{2}{x-2}+\dfrac{1}{x+2}=\dfrac{2x-5}{x^2-4}\)

Answer 1

\(\Leftrightarrow\dfrac{2\left(x+2\right)+x-2-2x+5}{\left(x-2\right)\left(x+2\right)}=0\left(ĐK:x\ne\pm2\right)\)

\(\Leftrightarrow2x+4+x-2-2x+5=0:\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x+7=0\)

\(\Leftrightarrow x=-7\left(n\right)\)

Vậy \(S=\left\{-7\right\}\)

Answer 2

\(=>\dfrac{2}{x-2}+\dfrac{1}{x+2}-\dfrac{2x-5}{x^2-4}=0\)

\(=>\dfrac{2\left(x+2\right)+x-2-2x+5}{\left(x+2\right)\left(x-2\right)}=0\)    ( điều kiện x ≠ ± 2)

\(=>2x+4+x+\left(-2\right)-2x+5=\dfrac{0}{\left(x+2\right)\left(x-2\right)}\)

\(=>x+7=0=>x=-7\left(n\right)\)