Question

[Nguyễn Hồng Nhung]

giải nhanh giúp mik đko ạ

Answer 1

[Xyz OLM]

ĐKXĐ : \(x\ne-1;x\ne2;x\ne-4\)

Ta có : \(\dfrac{2x+5}{x^2+5x+4}-\dfrac{2x+2}{\left(x-2\right).\left(x+4\right)}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x+1+x+4}{\left(x+1\right).\left(x+4\right)}-\dfrac{x-2+x+4}{\left(x-2\right).\left(x+4\right)}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{x+4}+\dfrac{1}{x+1}-\dfrac{1}{x+4}-\dfrac{1}{x-2}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x-2}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{-3}{x^2-x-2}=\dfrac{3}{2}\)

\(\Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\left(tm\right)\)

Tập nghiệm phương trình S = {0;1}