Question

giải phương trình \(\frac{x}{x^2+4x+4}\)+\(\frac{5x}{x^2+4}\)=-2

Answer 1

\(\frac{x}{x^2+4x+4}+\frac{5x}{x^2+4}=-2^{\left(1\right)}\)

\(ĐK:x\ne-2\)

\(\left(1\right)\Leftrightarrow\left(\frac{x}{x^2+4x+4}+1\right)+\left(\frac{5x}{x^2+4}+1\right)=0\)

\(\Leftrightarrow\frac{x^2+5x+4}{x^2+4x+4}+\frac{x^2+5x+4}{x^2+4}=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(\frac{1}{x^2+4x+4}+\frac{1}{x^2+4}\right)=0\)

\(\Leftrightarrow x^2+5x+4=0\left(vì\frac{1}{x^2+4x+4}+\frac{1}{x^2+4}>0\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}\left(t/mĐK\right)}\)

vậy pt đã cho có tập nghiệm S={-1;-4}