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

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1. `a) 5+76/(x^2-16)=(2x-1)/(x+4)-(3x-1)/(4-x)` (ĐK: `x \ne +-4`)

   `<=> (5(x^2-16)+76)/((x-4)(x+4))=((2x-1)(x-4)+(3x-1)(x+4))/((x-4)(x+4))`

   `=> 5x^2-80+76=2x^2-8x-x+4+3x^2+12x-x-4`

   `<=> 5x^2-4=5x^2+2x`

   `<=> 2x=-4`

   `<=> x=-2` ™

   Vậy `S={-2}`

   `b) 3/(x+2)-2/(x-2)+8/(x^2-4)=0` (ĐK: `x \ne +-2`)

   `<=> (3(x-2)-2(x+2)+8)/((x-2)(x+2))=0`

   `=> 3x-6-2x-4+8=0`

   `<=> x-2=0`

   `<=> x=2` (ktm)

   Vậy `S=∅`

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2. `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 phương trình vô nghiệm.

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