Question

Giải phương trình:

\(\dfrac{x}{x-3}\) + \(\dfrac{x}{x+2}\) = \(\dfrac{3x+6}{\left(x-3\right)\left(x+2\right)}\)

Answer 1

\(\dfrac{x}{x-3}+\dfrac{x}{x+2}=\dfrac{3x+6}{\left(x-3\right)\left(x+2\right)}\)    (1)

ĐKXĐ: \(x\ne3;x\ne-2\)

\(\left(1\right)\Leftrightarrow x\left(x+2\right)+x\left(x-3\right)=3x+6\)

\(\Leftrightarrow x^2+2x+x^2-3x=3x+6\)

\(\Leftrightarrow2x^2-4x-6=0\)

\(\Leftrightarrow2\left(x^2-2x-3\right)=0\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow x^2+x-3x-3=0\)

\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0;x-3=0\)

*) \(x+1=0\)

\(\Leftrightarrow x=-1\) (nhận)

*) \(x-3=0\)

\(\Leftrightarrow x=3\) (loại)

Vậy \(S=\left\{-1\right\}\)