Question

giải phương trình: \(\dfrac{4}{x+2}-\dfrac{1}{x}=\dfrac{x^2-2}{x^2+2x}\)

Answer 1

\(\dfrac{4}{x+2}-\dfrac{1}{x}=\dfrac{x^2-2}{x^2+2x}\left(x\ne0;x\ne-2\right)\)

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

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

\(\Rightarrow4x-\left(x+2\right)=x^2-2\)

\(\Leftrightarrow4x-x-x^2=2-2\)

\(\Leftrightarrow3x-x^2=0\)

\(\Leftrightarrow x\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=3\left(tm\right)\end{matrix}\right.\)