## Tìm X: (2x +1):(2x^2 + 5x + 2) – ( 3 : x^2 -4) = 2

Question

Tìm X:
(2x +1):(2x^2 + 5x + 2) – ( 3 : x^2 -4) = 2

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9 months 2021-05-10T11:08:15+00:00 2 Answers 7 views 0

1. 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

$$\left[ \begin{array}{l}x= \dfrac{3}{2}\\x=-1\end{array} \right.$$

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\}