Xin cách giải ạ Huhu giúp em với

ĐKXĐ: `x >= 1`

`sqrt(3x-2)+sqrt(x-1)=4x-9+2sqrt(3x^2-5x+2)`

`<=>` `sqrt(3x-2)+sqrt(x-1)=(3x-2)+(x-1)-6+2sqrt((3x-2)(x-1))`

`<=>` `sqrt(3x-2)+sqrt(x-1)=(sqrt(3x-2)+sqrt(x-1))^2-6`

`<=>` `(sqrt(3x-2)+sqrt(x-1))^2-(sqrt(3x-2)+sqrt(x-1))-6=0`

`=>` `sqrt(3x-2)+sqrt(x-1)=3`

`=>` `3x-2+x-1+2sqrt((3x-2)(x-1))=9`

`<=>` `2sqrt((3x-2)(x-1))=-4x+12`

`<=>` `sqrt(3x^2-5x+2)=-2x+6`

`=>` `{(3x^2-5x+2=4x^2-24x+36),(x <= 3):}`

`<=>` `{(x^2-19x+34=0),(x <= 3):}`

`<=>` `{(x=17 \ "hoặc" \ x = 2),(x <= 3):}`

`=>` `x=2` (Thỏa mãn)

Vậy `x=2`