Question

giải pt:

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

Answer 1

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{\left(x-2\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{5}{\left(x-2\right)\left(x+3\right)}\)

\(\Leftrightarrow x+3-6x+12=5\)

\(\Leftrightarrow-5x+15=5\)

\(\Leftrightarrow x=2\)