`4/{x-1}-5/{x-2}=-3` `ĐK: x≠1;x≠2 `
`⇔{4(x-2)}/{(x-1)(x-2)}-{5(x-1)}/{(x-1)(x-2)}={-3(x-1)(x-2)}/{(x-1)(x-2)}`
`⇔4x-8-5x+5=-3x²+9x-6`
`⇔3x²+4x-5x-9x+5-8+6=0`
`⇔3x²-10x+3=0`
`⇔3x²-9x-x+3=0`
`⇔3x(x-3)-(x-3)=0`
`⇔(x-3)(3x-1)=0`
⇒\(\left[ \begin{array}{l}x-3=0\\3x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3(tm)\\x=\dfrac{1}{3}(tm)\end{array} \right.\)
`Vậy` `x∈`{`3;\frac{1}{3}` }
