Question

Giải PT (nhớ tìm ĐKXĐ)

5+96/x²-16=2x-1/x+4+3x-1/x-4

 

Answer 1

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)ĐKXĐ : \(x\ne\pm4\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2-11x+9x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\)( t/m )

Vậy....