Question

Mọi người ơi giúp em với !!! Rút gọn biểu thức :

Q= \(\dfrac{2}{2x-1}\sqrt{8x^2\left(1-4x+4x^2\right)}\)với x>0 , x khac 1/2

Answer 1

\(Q=\dfrac{2}{2x-1}\sqrt{8x^2\left(1-4x+4x^2\right)}\)

\(=\dfrac{2}{2x-1}\cdot2x\sqrt{2\left(1-4x+4x^2\right)}\)

\(=\dfrac{4x\sqrt{2\left(1-4x+4x^2\right)}}{2x-1}\)

\(=\dfrac{4x\sqrt{2-8x+8x^2}}{2x-1}\)

\(=\dfrac{4x\sqrt{\left(\sqrt{2}-\sqrt{8}x\right)^2}}{2x-1}\)

\(=\dfrac{4x\cdot\left(\sqrt{2}-\sqrt{8}x\right)}{2x-1}\)

\(=\dfrac{4x\cdot\left(\sqrt{2}-2\sqrt{2}x\right)}{2x-1}\)

\(=\dfrac{4\sqrt{2}x-8\sqrt{2}x^2}{2x-1}\)

Answer 2

điều kiện \(x>\dfrac{1}{2}\)

Q = \(\dfrac{2}{2x-1}\sqrt{8x^2\left(1-4x+4x^2\right)}\) = \(\dfrac{2}{2x-1}\sqrt{\left(\sqrt{8}x\right)^2\left(2x-1\right)^2}\)

Q = \(\dfrac{2}{2x-1}\left(\sqrt{8}x\right)\left(2x-1\right)\) \(\left(vìx>\dfrac{1}{2}\right)\)

Q = \(2\sqrt{8}x\)