\(1+\frac{x}{x+3}-\frac{2}{x+2}=\frac{5x}{\left(x+2\right)\left(3-x\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne3\\x\ne-2\\x\ne-3\end{matrix}\right.\)
\(MSC:\left(x+2\right)\left(3-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x+2\right)\left(3-x\right)\left(x+3\right)-x\left(x+2\right)\left(3-x\right)-2\left(x+3\right)\left(3-x\right)=5x\left(3+x\right)\)
\(\Leftrightarrow x^2-2x^3+15x-5x^2-15x=0\)
\(\Leftrightarrow x^2-2x^3-5x^2=0\)
\(\Leftrightarrow x^2\left(2+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Vậy .............
Mình nghĩ đề là \(1+\frac{x}{x-3}-\frac{2}{x+2}=\frac{5x}{\left(x+2\right)\left(3-x\right)}\) thì đúng hơn á..bạn xem lại nha
\(1+\frac{x}{x-3}-\frac{2}{x+2}=\frac{5x}{\left(x+2\right)\left(3-x\right)}\)
\(\Leftrightarrow1-\frac{x}{3-x}+\frac{2}{x+2}=\frac{5x}{\left(x+2\right)\left(3-x\right)}\)
\(\Leftrightarrow\frac{1\left(x+2\right)\left(3-x\right)-x\left(x+2\right)-2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=\frac{5x}{\left(x+2\right)\left(3-x\right)}\)
\(\Leftrightarrow\frac{-x-12}{\left(x+2\right)\left(3-x\right)}-\frac{5x}{\left(x+2\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\frac{-6x-12}{\left(x+2\right)\left(3-x\right)}\)
\(\Leftrightarrow\frac{-6\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}\)
\(\Leftrightarrow\frac{-6}{3-x}\)