ĐKXĐ: \(\dfrac{x}{x+2}+\dfrac{2}{x-2}=\dfrac{4x}{4-x^2}\) (ĐK: \(x\ne2,x\ne-2\))
\(\Leftrightarrow\dfrac{x}{x+2}+\dfrac{2}{x-2}=-\dfrac{4x}{x^2-4}\)
\(\Leftrightarrow\dfrac{x}{x+2}+\dfrac{2}{x-2}=-\dfrac{4x}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow x\left(x-2\right)+2\left(x+2\right)=-4x\)
\(\Leftrightarrow x^2-2x+2x+4=-4x\)
\(\Leftrightarrow x^2+4=-4x\)
\(\Leftrightarrow x^2+4+4x=0\)
\(\Leftrightarrow x^2+2\cdot2\cdot x+2^2=0\)
\(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\left(ktmdk\right)\)
Vậy: \(x\in\varnothing\)
ĐKXĐ: x<>-2; x<>2
PT=>\(\dfrac{x^2-2x+2x+4}{\left(x+2\right)\left(x-2\right)}=\dfrac{-4x}{x^2-4}\)
=>x^2+4=-4x
=>x^2+4x+4=0
=>(x+2)^2=0
=>x+2=0
=>x=-2(loại)
