Giải phương trình này

`#Ju ly `

`(3x-1)/(x-1)-(2x+5)/(x+3)-8/(x^2+2x-3)=1(x ` $\ne$ `1;x ` $\ne$ `-3) `

`=>(3x-1)/(x-1)-(2x+5)/(x+3)-8/(x^2-x+3x-3)=1 `

`=>(3x-1)/(x-1)-(2x+5)/(x+3)-8/(x(x-1)+3(x-1))=1 `

`=>(3x-1)/(x-1)-(2x+5)/(x+3)-8/((x-1)(x+3))=1 `

`=>((3x -1)(x+3)-(x-1)(2x+5)-8)/((x-1)(x+3))=((x-1)(x+3))/((x-1)(x+3)) `

`=>3x^2+9x-x-3-2x^2-5x+2x+5-8=x^2+2x-3 `

`=>3x^2-2x^2-x^2+9x-x-5x+2x-2x=3-5+8-3 `

`=>3x=3 `

`=>x=1(KTM) `

Vậy pt vô `n_{o} `.