\(\left(\dfrac{2}{1+2x}+\dfrac{4x^2+1}{4x^2-1}-\dfrac{1}{1-2x}\right):\dfrac{2}{4x^2-1}\)
\(=\left(\dfrac{2\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}+\dfrac{-\left(4x^2+1\right)}{\left(1-2x\right)\left(1+2x\right)}-\dfrac{1\left(1+2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\cdot\dfrac{4x^2-1}{2}\)
\(=\left(\dfrac{2-4x-4x^2-1-1-2x}{\left(1+2x\right)\left(1-2x\right)}\right)\cdot\dfrac{\left(1-2x\right)\left(1+2x\right)}{-2}\)
\(=\left(\dfrac{-4x^2-6x}{\left(1+2x\right)\left(1-2x\right)}\right)\cdot\dfrac{\left(1-2x\right)\left(1+2x\right)}{-2}\)
\(=\dfrac{-2x\left(2x+3\right)\left(1-2x\right)\left(1+2x\right)}{\left(1+2x\right)\left(1-2x\right)\cdot\left(-2\right)}\)
\(=\dfrac{x\left(2x+3\right)}{1}\)
\(=x\left(2x+3\right)\)
Để A = 2 thì \(x\left(2x+3\right)=2=1\cdot2=2\cdot1=\left(-1\right)\cdot\left(-2\right)=\left(-2\right)\cdot\left(-1\right)\)
Ta có bảng :
| x | 1 | 2 | -1 | -2 |
| 2x+3 | 2 | 1 | -2 | -1 |
| x1 | 1 | 2 | -1 | -2 |
| x2 | -0,5 | -1 | -2,5 | -2 |
Ta thấy chỉ có x = -2 và 2x + 3 = -1 thì x1 và x2 mới bằng nhau và bằng -2
Vậy x = -2 thì A = 2