Question

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

giải pt nhen

Answer 1

\(\dfrac{2x+1}{x+3}-\dfrac{x+3}{x-3}=\dfrac{x+1}{x^2-9}+3\)(ĐKXĐ: \(x\notin\left\{3;-3\right\}\))

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

=>\(\dfrac{\left(2x+1\right)\left(x-3\right)-\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x^2+x-26}{\left(x-3\right)\left(x+3\right)}\)

=>\(2x^2-6x+x-3-x^2-6x-9=3x^2+x-26\)

=>\(3x^2+x-26=x^2-11x-12\)

=>\(2x^2+12x-14=0\)

=>\(x^2+6x-7=0\)

=>(x+7)(x-1)=0

=>\(\left[{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=1\left(nhận\right)\end{matrix}\right.\)