\(\dfrac{x+2}{x}-\dfrac{x^2+5x+4}{x\left(x+2\right)}=\dfrac{x}{x+2}\)
\(ĐK:x\ne0;x\ne-2\)
\(\Leftrightarrow\dfrac{x^2+5x+4}{x\left(x+2\right)}=\dfrac{x^2}{x\left(x+2\right)}-\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}\)
\(\Leftrightarrow x^2+5x+4=x^2-\left(x+2\right)^2\)
\(\Leftrightarrow x^2+5x+4=x^2-\left(x^2+4x+4\right)\)
\(\Leftrightarrow x^2+5x+4=-4x-4\)
\(\Leftrightarrow x^2+9x+8=0\)
\(\Leftrightarrow x^2+x+8x+8=0\)
\(\Leftrightarrow x\left(x+1\right)+8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-8\end{matrix}\right.\) ( tm )
Vậy \(S=\left\{-1;-8\right\}\)
( ở trên sai nha )
\(\dfrac{x+2}{x}-\dfrac{x^2+5x+4}{x\left(x+2\right)}=\dfrac{x}{x+2}\)
\(ĐK:x\ne0;-2\)
\(\Leftrightarrow\dfrac{x+2}{x}-\dfrac{x}{x+2}=\dfrac{x^2+5x+4}{x\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}-\dfrac{x^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}\)
\(\Leftrightarrow\left(x^2+4x+4\right)-x^2=x^2+5x+4\)
\(\Leftrightarrow4x+4=x^2+5x+4\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
