Question

\(5+\dfrac{96}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)

Answer 1

\(\Leftrightarrow5x^2-80+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2+16=2x^2-8x-x+4+3x^2+12x-x-4\)

\(\Leftrightarrow5x^2+16=5x^2+2x\)

=>x=8(nhận)

Answer 2

đk x ≠ 4; x ≠ -4

\(\dfrac{5\left(x^2-16\right)+96}{x^2-16}=\dfrac{\left(2x-1\right)}{x+4}+\dfrac{3x-1}{x-4}\\ \dfrac{5x^2-80+96}{x^2-16}=\dfrac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{x^2-16}\\ 5x^2+16=2x^2-x-8x+4+3x^2-x+12x-4\\ 5x^2+16-5x^2+2x=0\\2x+16=0\\ x=-8\left(tho\text{a}m\text{a}nđk\right) \)