x-1/x+1/x+1=2x-1/x^2+x c,x+1/x-1-x-1/x+1=16/x^2-1 d,1+x/3-x=5x/(x+2)(3-x)+2/x+2 02/10/2021 Bởi Daisy x-1/x+1/x+1=2x-1/x^2+x c,x+1/x-1-x-1/x+1=16/x^2-1 d,1+x/3-x=5x/(x+2)(3-x)+2/x+2
$#Dino$ `c) (x+1)/(x-1)-(x-1)/(x+1)=16/(x²-1)` `ĐKXĐ: x`$\neq$ `±1` `⇔((x+1)²)/(x²-1)-((x-1)²)/(x²-1)=16/(x²-1)` `⇔(x+1)²-(x-1)²=16` `⇔x²+2x+1-x²+2x-1=16` `⇔4x=16` `⇔x=4` Vậy `S={4}` `d)(1+x)/(3-x)=(5x)/((x+2)(3-x))+2/(x+2)` `ĐKXĐ: x`$\neq$ `3;x`$\neq$ `-2` `⇔((x+2)(1+x))/((x+2)(3-x))=(5x)/((x+2)(3-x))+(2(3-x))/((x+2)(3-x))` `⇒(x+2)(1+x)=5x+2(3-x)` `⇔x+x²+2+2x=5x+6-2x` `⇔x²+2x+2x+x-5x-6+2=0` `⇔x²-4=0` `⇔(x-2)(x+2)=0` `1)x-2=0⇔x=2` `2)x+2=0⇔x=-2` Vậy `S={±2}` Bình luận
$#Dino$
`c) (x+1)/(x-1)-(x-1)/(x+1)=16/(x²-1)`
`ĐKXĐ: x`$\neq$ `±1`
`⇔((x+1)²)/(x²-1)-((x-1)²)/(x²-1)=16/(x²-1)`
`⇔(x+1)²-(x-1)²=16`
`⇔x²+2x+1-x²+2x-1=16`
`⇔4x=16`
`⇔x=4`
Vậy `S={4}`
`d)(1+x)/(3-x)=(5x)/((x+2)(3-x))+2/(x+2)`
`ĐKXĐ: x`$\neq$ `3;x`$\neq$ `-2`
`⇔((x+2)(1+x))/((x+2)(3-x))=(5x)/((x+2)(3-x))+(2(3-x))/((x+2)(3-x))`
`⇒(x+2)(1+x)=5x+2(3-x)`
`⇔x+x²+2+2x=5x+6-2x`
`⇔x²+2x+2x+x-5x-6+2=0`
`⇔x²-4=0`
`⇔(x-2)(x+2)=0`
`1)x-2=0⇔x=2`
`2)x+2=0⇔x=-2`
Vậy `S={±2}`