Question

\(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)

Giúp mk với

Answer 1

đk: x khác 2;4

\(\dfrac{2}{\left(x-4\right)\left(2-x\right)}-\dfrac{x-1}{x-2}-\dfrac{x+3}{x-4}=0\)

<=> \(\dfrac{-2-\left(x-1\right)\left(x-4\right)-\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)

<=> \(-2-x^2+5x-4-x^2-x+6=0\)

<=> -2x² +4x = 0

<=> 2x(x-2) = 0

Mà x khác 2

<=> 2x = 0

<=> x = 0