A=(2/1+2x+4xmu2+1/4xmu2-1-1/1-2x):2/4xmu2 -1 A rut gon bieu thuc A B tim x de A=2

Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne \dfrac{1}{2};x \ne  - \dfrac{1}{2}\\
A = \left( {\dfrac{2}{{1 + 2x}} + \dfrac{{4{x^2} + 1}}{{4{x^2} - 1}} - \dfrac{1}{{1 - 2x}}} \right):\dfrac{2}{{4{x^2} - 1}}\\
 = \left( {\dfrac{2}{{1 + 2x}} + \dfrac{{4{x^2} + 1}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} + \dfrac{1}{{2x - 1}}} \right).\dfrac{{\left( {2x + 1} \right)\left( {2x - 1} \right)}}{2}\\
 = \dfrac{{2\left( {2x - 1} \right) + 4{x^2} + 1 + 2x + 1}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}.\dfrac{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}{2}\\
 = \dfrac{{4x - 2 + 4{x^2} + 2x + 2}}{2}\\
 = 2{x^2} + 3\\
b)A = 2\\
 \Rightarrow 2{x^2} + 3 = 2\\
 \Rightarrow 2{x^2} + 1 = 0\left( {vn} \right)
\end{array}$

Vậy ko có giá trị của x để A=2