Question

Giải phương trình: \(\dfrac{x+1}{x-3}+\dfrac{x+1}{x+3}=\dfrac{2\left(x^2+5\right)}{x^2-9}\)

Answer 1

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\dfrac{x+1}{x-3}+\dfrac{x+1}{x+3}=\dfrac{2\left(x^2+5\right)}{x^2-9}\)

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

=>\(\dfrac{x^2+4x+3+x^2-2x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x^2+10}{\left(x-3\right)\left(x+3\right)}\)

=>\(2x^2+2x=2x^2+10\)

=>2x=10

=>x=5(nhận)