\(\dfrac{1}{x+2}+\dfrac{1}{x-3}=\dfrac{10}{x^2-x-6}\left(đlxđ:x\ne-2;x\ne3\right)\)
\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{1}{x-3}=\dfrac{10}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow x-3+x+2=10\)
\(\Leftrightarrow x+x=10+3-2\)
\(\Leftrightarrow2x=11\)
\(\Leftrightarrow x=\dfrac{11}{2}\left(nhận\right)\)
Vậy \(S=\left\{\dfrac{11}{2}\right\}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)
\(\dfrac{1}{x+2}+\dfrac{3}{x-3}=\dfrac{10}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{x-3}{\left(x+2\right)\left(x-3\right)}+\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{10}{x^2+2x-3x-6}\)
\(\Leftrightarrow\dfrac{x-3}{\left(x+2\right)\left(x-3\right)}+\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{10}{\left(x+2\right)\left(x-3\right)}\)
\(\Rightarrow x-3+3x+6=10\)
\(\Leftrightarrow4x=10-3\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
Vậy...
đkxđ : x khác 3
\(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)
\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{3}{x-3}=\dfrac{10}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{x-3}{\left(x+2\right)\left(x-3\right)}+\dfrac{3x+6}{\left(x-3\right)\left(x+2\right)}-\dfrac{10}{\left(x+2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x-3+3x+6-10=0\)
\(\Leftrightarrow4x-7=0\)
\(x=\dfrac{7}{4}\)( nhận)
