Toán Tìm X: (2x +1):(2x^2 + 5x + 2) – ( 3 : x^2 -4) = 2 08/09/2021 By Parker Tìm X: (2x +1):(2x^2 + 5x + 2) – ( 3 : x^2 -4) = 2
Đáp án: `S=\{-1;3/2\}` Giải thích các bước giải: `ĐKXĐ:x\ne -1/2;x\ne±2` `(2x+1):(2x^2+5x+2)-3:(x^2-4)=2` `⇔(2x+1)/(2x^2+5x+2)-3/(x^2-4)=2` `⇔(2x+1)/(2x^2+x+4x+2)-3/(x^2-4)=2` `⇔(2x+1)/[x(2x+1)+2(2x+1)]-3/(x^2-4)=2` `⇔(2x+1)/((2x+1)(x+2))-3/(x^2-4)=2` `⇔1/(x+2)-3/(x^2-4)=2` `⇔(x-2)/((x-2)(x+2))-3/((x-2)(x+2))=(2(x^2-4))/((x-2)(x+2))` `⇒(x-2)-3=2(x^2-4)` `⇔x-5=2x^2-8` `⇔2x^2-8-x+5=0` `⇔2x^2-x-3=0` `⇔2x^2+2x-3x-3=0` `⇔2x(x+1)-3(x+1)=0` `⇔(x+1)(2x-3)=0` \(⇔\left[ \begin{array}{l}x+1=0\\2x-3=0\end{array} \right.\) \(\left[ \begin{array}{l}x=-1(TM)\\x=\dfrac{3}{2}(TM)\end{array} \right.\) Vậy `S=\{-1;3/2\}` Trả lời
Đáp án: Giải thích các bước giải: $\dfrac{\left(2x+1\right)}{2x^2+5x+2}-\:\dfrac{3}{x^2-4}=2$ `<=>` $\dfrac{1}{x+2}-\dfrac{3}{x^2-4}=2$ `<=>` $x-5=2\left(x+2\right)\left(x-2\right)$ `<=>2x^2-8=x-5` `<=>2x^2-x-3=0` `<=>(2x-3)(x+1)=0` \(\left[ \begin{array}{l}x= \dfrac{3}{2}\\x=-1\end{array} \right.\) Trả lời
Đáp án:
`S=\{-1;3/2\}`
Giải thích các bước giải:
`ĐKXĐ:x\ne -1/2;x\ne±2`
`(2x+1):(2x^2+5x+2)-3:(x^2-4)=2`
`⇔(2x+1)/(2x^2+5x+2)-3/(x^2-4)=2`
`⇔(2x+1)/(2x^2+x+4x+2)-3/(x^2-4)=2`
`⇔(2x+1)/[x(2x+1)+2(2x+1)]-3/(x^2-4)=2`
`⇔(2x+1)/((2x+1)(x+2))-3/(x^2-4)=2`
`⇔1/(x+2)-3/(x^2-4)=2`
`⇔(x-2)/((x-2)(x+2))-3/((x-2)(x+2))=(2(x^2-4))/((x-2)(x+2))`
`⇒(x-2)-3=2(x^2-4)`
`⇔x-5=2x^2-8`
`⇔2x^2-8-x+5=0`
`⇔2x^2-x-3=0`
`⇔2x^2+2x-3x-3=0`
`⇔2x(x+1)-3(x+1)=0`
`⇔(x+1)(2x-3)=0`
\(⇔\left[ \begin{array}{l}x+1=0\\2x-3=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1(TM)\\x=\dfrac{3}{2}(TM)\end{array} \right.\)
Vậy `S=\{-1;3/2\}`
Đáp án:
Giải thích các bước giải:
$\dfrac{\left(2x+1\right)}{2x^2+5x+2}-\:\dfrac{3}{x^2-4}=2$
`<=>` $\dfrac{1}{x+2}-\dfrac{3}{x^2-4}=2$
`<=>` $x-5=2\left(x+2\right)\left(x-2\right)$
`<=>2x^2-8=x-5`
`<=>2x^2-x-3=0`
`<=>(2x-3)(x+1)=0`
\(\left[ \begin{array}{l}x= \dfrac{3}{2}\\x=-1\end{array} \right.\)