ĐKXĐ: \(x\notin\left\{4;-4\right\}\)
\(\dfrac{2x-5}{x+4}+\dfrac{x}{4-x}=\dfrac{17x-56}{16-x^2}\)
=>\(\dfrac{2x-5}{x+4}-\dfrac{x}{x-4}=\dfrac{-17x+56}{\left(x-4\right)\left(x+4\right)}\)
=>\(\dfrac{\left(2x-5\right)\left(x-4\right)-x\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{-17x+56}{\left(x-4\right)\left(x+4\right)}\)
=>\(2x^2-8x-5x+20-x^2-4x=-17x+56\)
=>\(x^2-17x+20+17x-56=0\)
=>\(x^2-36=0\)
=>\(x^2=36\)
=>\(\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
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