a. 5+76/x^2-16=2x-1/x+4-3x -1/4-x b. 3/x+2-2/x-2+8/x^2-4=0

Đáp án:

`a)5+76/(x²-16)=(2x-1)/(x+4)-(3x-1)/(4-x)(` `ĐKXĐ:“x`$\neq$`±4)`

`⇔5+76/(x²-16)=(2x-1)/(x+4)+(3x-1)/(x-4)`

`⇔[5(x²-16)]/(x²-16)+76/(x²-16)=[(2x-1)(x-4)]/(x²-16)+[(3x-1)(x+4)]/(x²-16)`

`⇒5(x²-16)+76=(2x-1)(x-4)+(3x-1)(x+4)`

`⇔5x²-80+76=2x²-8x-x+4+3x²+12x-x-4`

`⇔5x²-2x²+8x+x-3x²-12x+x=4-4+80-76`

`⇔-2x=4`

`⇔x=-2(TM` `ĐKXĐ)`

Vậy `S={-2}`

`b)3/(x+2)-2/(x-2)+8/(x²-4)=0(` `ĐKXĐ:“x`$\neq$ `±2)`

`⇔[3(x-2)]/(x²-4)-[2(x+2)]/(x²-4)+8/(x²-4)=0`

`⇒3(x-2)-2(x+2)+8=0`

`⇔3x-6-2x-4+8=0`

`⇔x-2=0`

`⇔x=2(Ko` `TM` `ĐKXĐ)`

Vậy `S=\emptyset`