giúp tớ làm bài tập nha

Đáp án+Giải thích các bước giải:

 Bài `12:`

`a)`

Ta có:

`2x^{2}+6x=2x.(x+3);`

`x^{2}-9=x^{2}-3^{2}=(x-3)..(x+3)`

`->MTC=2x.(x-3).(x+3)`

Ta lại có:

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

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

Ta quy đồng:

`\frac{7x-1}{2x^{2}+6x}=\frac{(7x-1).(x-3)}{2x.(x-3).(x+3)}=\frac{7x^{2}-22x+3}{2x.(x-3).(x+3)};`

`\frac{5-3x}{x^{2}-9}=\frac{(5-3x).2x}{2x.(x-3).(x+3)}=\frac{10x-6x^{2}}{2x.(x-3).(x+3)}`

Vậy `\frac{7x-1}{2x^{2}+6x}=\frac{7x^{2}-22x+3}{2x.(x-3).(x+3)};`

       `\frac{5-3x}{x^{2}-9}=\frac{10x-6x^{2}}{2x.(x-3).(x+3)}.`

`b)`

Ta xét:

`x-x^{2}=x.(1-x);`

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

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

Lại có:

`2x.(1-x)^{2}:x.(1-x)=2.(1-x)` ;

`2x.(1-x)^{2}:2.(1-x)^{2}=x`

Ta quy đồng:

`\frac{x+1}{x-x^{2}}=\frac{(x+1).2.(1-x)}{2x.(1-x)^{2}}=\frac{2.(1-x^{2})}{2x.(1-x)^{2}}=\frac{2-2x^{2}}{2x.(1-x)^{2}}` ;

`\frac{x+2}{2-4x+2x^{2}}=\frac{(x+2).x}{2x.(1-x)^{2}}=\frac{x^{2}+2x}{2x.(1-x)^{2}}`

Vậy `\frac{x+1}{x-x^{2}}=\frac{2-2x^{2}}{2x.(1-x)^{2}};`

`\frac{x+2}{2-4x+2x^{2}}=\frac{x^{2}+2x}{2x.(1-x)^{2}}`

Bài `13:`

`a)`

`\frac{x+y}{xy}+\frac{x-y}{xy}` `(x\ne0;y\ne0)`

`=\frac{x+y+x-y}{xy}`

`=\frac{2x}{xy}`

`=2/y`

`b)`

`\frac{x^{2}+5x}{x+2}-\frac{x-4}{x+2}`

`(đk:x\ne-2)`

`=\frac{x^{2}+5x-x+4}{x+2}`

`=\frac{x^{2}+2x+2x+4}{x+2}`

`=\frac{x.(x+2)+2.(x+2)}{x+2}`

`=\frac{(x+2).(x+2)}{x+2}`

`=x+2`

`c)`

`\frac{3x+2y}{x^{2}-y^{2}}-\frac{x}{x^{2}-y^{2}}`

`(đk:x\ne\pm y)`

`=\frac{3x+2y-x}{x^{2}-y^{2}}`

`=\frac{2x+2y}{(x-y).(x+y)}`

`=\frac{2.(x+y)}{(x-y).(x+y)}`

`=\frac{2}{x-y}`