Đáp án+Giải thích các bước giải:

 $(*)$

`\frac{x^{2}-2x-8}{2x^{2}-8}=\frac{x-4}{A}`

`(đk:x\ne\pm2;A\ne0)`

`->A.(x^{2}-4x+2x-8)=[2.(x^{2}-2^{2})].(x-4)`

`->A=\frac{2.(x-2).(x+2).(x-4)}{x.(x-4)+2.(x-4)}`

`->A=\frac{2.(x-2).(x+2).(x-4)}{(x+2).(x-4)}`

`->A=2.(x-2)`

`->A=2x-4`

Vậy `A=2x-4`

$(*)$ `\frac{A}{7x^{2}+9x+2}=\frac{4x-3}{7x+2}`

`(đk:x\ne -2/7;x\ne-1)`

`->A.(7x+2)=(4x-3).(7x^{2}+7x+2x+2)`

`->A=\frac{(4x-3).[7x.(x+1)+2.(x+1)]}{7x+2}`

`->A=\frac{(4x-3).(7x+2).(x+1)}{7x+2}`

`->A=(4x-3).(x+1)`

`->A=4x^{2}+4x-3x-3`

`->A=4x^{2}+x-3`

Vậy `A=4x^{2}+x-3.`