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 14/07/2021 Bởi Eliza 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` Bình luận
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`