$\frac{x+1}{x-2}$ -$\frac{x-1}{x+2}$ =$\frac{2(x^2+2)}{x^2-4}$ 08/10/2021 Bởi aikhanh $\frac{x+1}{x-2}$ -$\frac{x-1}{x+2}$ =$\frac{2(x^2+2)}{x^2-4}$
`(x+1)/(x-2)-(x-1)/(x+2)=(2(x^2+2))/(x^2-4)` ĐKXĐ: `x\ne2;x\ne-2` `<=>frac{(x+1)(x+2)}{(x-2)(x+2)}-frac{(x-1)(x-2)}{(x+2)(x-2)}=frac{2x^2+4}{(x-2)(x+2)}` `=>(x+1)(x+2)-(x-1)(x-2)=2x^2+4` `<=>x^2+2x+x+2-(x^2-2x-x+2)=2x^2+4` `<=>x^2+3x+2-x^2+3x-2=2x^2+4` `<=>6x=2x^2+4` `<=>2x^2-6x+4=0` `<=>2x^2-2x-4x+4=0` `<=>(2x^2-2x)-(4x-4)=0` `<=>2x(x-1)-4(x-1)=0` `<=>(x-1)(2x-4)=0` `<=>` \(\left[ \begin{array}{l}x-1=0\\2x-4=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\2x=4\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1(TMĐK)\\x=2(KTMĐK)\end{array} \right.\) Vậy phương trình trên có nghiệm `S={1}` Bình luận
Đáp án: `S={1}` Giải thích các bước giải: ĐKXĐ: \(\begin{cases}x\neq2\\x\neq-2\\\end{cases}\) \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2(x^2+2)}{x^2-4}\\\Leftrightarrow \dfrac{(x+1)(x+2)-(x-1)(x-2)}{(x-2)(x+2)}=\dfrac{2x^2+4}{(x-2)(x+2)}\\\Rightarrow (x+1)(x+2)-(x-1)(x-2)=2x^2+4\\\Leftrightarrow x^2+3x+2-(x^2-3x+2)=2x^2+4\\\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\\\Leftrightarrow -2x^2+6x-4=0\\\Leftrightarrow -2x^2+2x+4x-4=0\\\Leftrightarrow -2x(x-1)+4(x-1)=0\\\Leftrightarrow (x-1)(-2x+4)=0\\\Leftrightarrow \left[ \begin{array}{l}x-1=0\\-2x+4=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1\\-2x=-4\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1\,\,(tmđk)\\x=2\,\,(ktmđk)\end{array} \right.\) Vậy `S={1}` Bình luận
`(x+1)/(x-2)-(x-1)/(x+2)=(2(x^2+2))/(x^2-4)` ĐKXĐ: `x\ne2;x\ne-2`
`<=>frac{(x+1)(x+2)}{(x-2)(x+2)}-frac{(x-1)(x-2)}{(x+2)(x-2)}=frac{2x^2+4}{(x-2)(x+2)}`
`=>(x+1)(x+2)-(x-1)(x-2)=2x^2+4`
`<=>x^2+2x+x+2-(x^2-2x-x+2)=2x^2+4`
`<=>x^2+3x+2-x^2+3x-2=2x^2+4`
`<=>6x=2x^2+4`
`<=>2x^2-6x+4=0`
`<=>2x^2-2x-4x+4=0`
`<=>(2x^2-2x)-(4x-4)=0`
`<=>2x(x-1)-4(x-1)=0`
`<=>(x-1)(2x-4)=0`
`<=>` \(\left[ \begin{array}{l}x-1=0\\2x-4=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\2x=4\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1(TMĐK)\\x=2(KTMĐK)\end{array} \right.\)
Vậy phương trình trên có nghiệm `S={1}`
Đáp án:
`S={1}`
Giải thích các bước giải:
ĐKXĐ: \(\begin{cases}x\neq2\\x\neq-2\\\end{cases}\)
\(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2(x^2+2)}{x^2-4}\\\Leftrightarrow \dfrac{(x+1)(x+2)-(x-1)(x-2)}{(x-2)(x+2)}=\dfrac{2x^2+4}{(x-2)(x+2)}\\\Rightarrow (x+1)(x+2)-(x-1)(x-2)=2x^2+4\\\Leftrightarrow x^2+3x+2-(x^2-3x+2)=2x^2+4\\\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\\\Leftrightarrow -2x^2+6x-4=0\\\Leftrightarrow -2x^2+2x+4x-4=0\\\Leftrightarrow -2x(x-1)+4(x-1)=0\\\Leftrightarrow (x-1)(-2x+4)=0\\\Leftrightarrow \left[ \begin{array}{l}x-1=0\\-2x+4=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1\\-2x=-4\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1\,\,(tmđk)\\x=2\,\,(ktmđk)\end{array} \right.\)
Vậy `S={1}`