Đáp án+Giải thích các bước giải:

 Bài `10:`

`a)(4x-8)/(2x^{2}+1)=0`

`(4x-8)/(2x^{2}+1)=(0)/(2x^{2}+1)`

`=>4x-8=0`

`4x=8`

`x=2`

Vậy `S={2}`

`b)(x^{2}-x-6)/(x-3)=0`

`( đ k : x\ne3)`

`(x^{2}-3x+2x-6)/(x-3)=0`

`(x.(x-3)+2.(x-3))/(x-3)=0`

`((x+2).(x-3))/(x-3)=(0)/(x-3)`

`(x+2).(x-3)=0`

`x+2=0` hoặc `x-3=0`

`x=-2(tmđk)` hoặc `x=3(ktmđk)`

Vậy `S={-2}`

`c)(x+5)/(3x-6)-1/2=(2x-3)/(2x-4)`

`( đ k : x\ne2)`

`(2.(x+5))/(6.(x-2))-(3.(x-2))/(6.(x-2))=(3.(2x-3))/(6.(x-2))`

`=>2x+10-3x+6=6x-9`

`-x+16-6x+9=0`

`-7x+25=0`

`-7x=-25`

`x=25/7(tmđk)`

Vậy `S={25/7}`

`d)(12)/(1-9x^{2})=(1-3x)/(1+3x)-(1+3x)/(1-3x)`

`( đ k : x\ne\pm 1/3)`

`(12)/((1-3x).(1+3x))=((1-3x)^{2})/((1-3x).(1+3x))-((1+3x)^{2})/((1-3x).(1+3x))`

`=>12=(1-3x)^{2}-(1+3x)^{2}`

`(1-3x-1-3x).(1-3x+1+3x)=12`

`(-6x).2=12`

`-6x=6`

`x=-1(tmđk)`

Vậy `S={-1}`

Câu `11:`

`a)5+(96)/(x^{2}-16)=(2x-1)/(x+4)-(3x-1)/(4-x)`

`( đ k : x\ne\pm4)`

`(5.(x-4).(x+4))/((x-4).(x+4))+(96)/((x-4).(x+4))=((2x-1).(x-4))/((x-4).(x+4))+((3x-1).(x+4))/((x-4).(x+4))`

`=>5.(x^{2}-4^{2})+96=2x^{2}-8x-x+4+3x^{2}+12x-x-4`

`5x^{2}-80+96=5x^{2}+2x`

`5x^{2}-5x^{2}+16=2x`

`2x=16`

`x=8(tmđk)`

Vậy `S={8}`

`b)(x+1)/(x-2)-(5)/(x+2)=(12)/(x^{2}-4)+1`

`( đ k : x\ne\pm2)`

`((x+1).(x+2))/((x-2).(x+2))-(5.(x-2))/((x-2).(x+2))=(12)/((x-2).(x+2))+((x-2).(x+2))/((x-2).(x+2))`

`=>x^{2}+2x+x+2-5x+10=12+(x^{2}-2^{2})`

`x^{2}-2x+12-12-x^{2}+4=0`

`-2x+4=0`

`-2x=-4`

`x=2(ktmđk)`

Vậy `S=\emptyset`

hay phương trình vô nghiệm

`c)(x+1)/(x^{2}+x+1)-(x-1)/(x^{2}-x+1)=(3)/(x.(x^{4}+x^{2}+1))`

`(x.(x+1).(x^{2}-x+1))/(x.(x^{2}-x+1).(x^{2}+x+1))-(x.(x-1).(x^{2}+x+1))/(x.(x^{2}-x+1).(x^{2}+x+1))=(3)/(x.(x^{2}-x+1).(x^{2}+x+1))`

`=>x.(x+1).(x^{2}-x.1+1^{2})-x.(x-1).(x^{2}+x.1+1^{2})=3`

`x.[x^{3}+1^{3}-(x^{3}-1^{3})]=3`

`x.(x^{3}+1-x^{3}+1)=3`

`x.2=3`

`x=3/2(tmđk)`

vậy `S={3/2}`