ý 28, 31 và ý 1, 4, 7

`28)`

`9x^3+30x^2+25x=0`

`<=>x.(9x^2+30x+25)=0`

`<=>x.[(3x)^2+2.3x.5+5^2]=0`

`<=>x.(3x+5)^2=0`

`<=>[(x=0),((3x+5)^2=0):}`

`<=>[(x=0),(3x+5=0):}`

`<=>[(x=0),(x=-5/3):}`

Vậy `S={0;-5/3}`

`31)`

`(2x-5).(9x^2-10x+1)=0`

`<=>(2x-5).(9x^2-9x-x+1)=0`

`<=>(2x-5).[9x.(x-1)-1.(x-1)]=0`

`<=>(2x-5).(9x-1).(x-1)=0`

`<=>[(x=5/2),(x=1/9),(x=1):}`

Vậy `S={5/2;1/9;1}`

Bài `12:`

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

ĐKXĐ : `xne-1` và `xne2`

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

`=>x^2-x-2-5x=2.(x-2)`

`<=>x^2-6x-2-2x+4=0`

`<=>x^2-8x+2=0`

`<=>x^2-8x+16=14`

`<=>(x-4)^2=14`

`<=>[(x=4+sqrt14),(x=4-sqrt14):}`

Vậy `S={4+sqrt14;4-sqrt14}`

`4)`

`1/(x+3)+1/(x-1)=1/x`

Đk : `xne3` `;` `xne1` `;` `xne0`

`<=>(x.(x-1)+x.(x+3))/(x.(x-1).(x+3))=((x+3).(x-1))/(x.(x-1).(x+3))`

`=>x^2-x+x^2+3x=x^2-x+3x-3`

`<=>2x^2+2x-x^2-2x+3=0`

`<=>x^2+3=0` (Vô lí)

Vậy phương trình vô nghiệm

`7)`

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

ĐKXĐ : `xne4`

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

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

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

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

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

`<=>-x^2+3x=0`

`<=>-x.(x-3)=0`

`<=>[(-x=0),(x-3=0):}`

`<=>[(x=0),(x=3):}` (thỏa mãn)

Vậy `S={0;3}`