Question

Answer 1

Ta có:

`(x+1)/(x-2) . (x-2)/(x-4)=(x+1)/(x-4)`

Đặt `a=(x+1)/(x-2);b=(x-2)/(x-4) (a,b ne0)`

`=>a.b=(x+1)/(x-4)`

`=>`Phương trình trở thành:

`a^2+ab-12b^2=0`

`<=>a^2+4ab-3ab-12b^2=0`

`<=>(a+4b)(a-3b)=0`

`<=>[(a=-4b),(a=3b):}`

Th1:

`a=-4b`

`=> (x+1)/(x-2)=-4. (x-2)/(x-4)`

`=>(x+1)(x-4)=-4(x-2)^2`

`<=>x^2+x-4x-4+4(x^2-4x+4)=0`

`<=>5x^2-19x+12=0`

`<=>[(x=3),(x=4/5):}(tm)`

th2:

`a=3b`

`=>(x+1)/(x-2)=3(x-2)/(x-4)`

`=>(x+1)(x-4)=3(x-2)^2`

`<=>x^2-3x-4=3x^2-12x+12`

`<=>2x^2-9x+16=0`(vô lý)

Vậy `x in {4/5;3}`