Question

Giải phương trình

\(\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+3}+\frac{1}{x+4}\)

Answer 1

\(\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+3}+\frac{1}{x+4}\)

ĐKXĐ : \(x\ne-2;x\ne-3;x\ne-4;x\ne-5\)

\(\Leftrightarrow\) \(\frac{2x+7}{\left(x+2\right)\left(x+5\right)}=\frac{2x+7}{\left(x+3\right)\left(x+4\right)}\)

\(\Leftrightarrow\) \(\left(2x+7\right)\left(\frac{1}{\left(x+2\right)\left(x+5\right)}-\frac{1}{\left(x+3\right)\left(x+4\right)}\right)=0\)

\(\Leftrightarrow\left(2x+7\right)\left(\frac{2}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)}\right)=0\)

\(\Rightarrow2x+7=0\Leftrightarrow x=-3,5\left(TM\right)\)

Vậy phương trình có tập nghiệm là \(S=\left\{-3,5\right\}\)