\(=\dfrac{2\sqrt{5}\left|a\left(2a-1\right)\right|}{2a-1}=\dfrac{2a\left(2a-1\right)\sqrt{5}}{2a-1}=2a\sqrt{5}\)

\(=\dfrac{2\sqrt{5}\cdot a\left(2a-1\right)}{2a-1}=2a\sqrt{5}\)

Q = (1 - \(\dfrac{\sqrt{a}-4a}{1-4a}\)_{) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\)}

     _{= \(\left(\dfrac{1-4a-\sqrt{a}+4a}{1-4a}\right):\left[\dfrac{1-4a-1-2a+4a+2\sqrt{a}}{1-4a}\right]\)}

    = \(\dfrac{1-\sqrt{a}}{1-4a}:\left(\dfrac{-2a+2\sqrt{a}}{1-4a}\right)\)

    = \(\dfrac{1-\sqrt{a}}{1-4a}.\dfrac{1-4a}{2\sqrt{a}\left(1-\sqrt{a}\right)}\)

    = \(\dfrac{1}{2\sqrt{a}}\) = \(\dfrac{\sqrt{a}}{2a}\)

\(B=\frac{1}{2a-1}.\sqrt{5a^4\left(2a-1\right)^2}=\sqrt{5}a^2.\frac{\left|2a-1\right|}{2a-1}\)

Nếu \(a>\frac{1}{2}\) thì \(B=\sqrt{5}a^2\)

Nếu \(a< \frac{1}{2}\) thì \(B=-\sqrt{5}a^2\)

\(B=\frac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\)

\(B=\frac{2\left|a\right|}{2a-1}\sqrt{5\left[1-2.2a+\left(2a\right)^2\right]}\)

\(B=\frac{2a}{2a-1}\sqrt{5\left(1-2a\right)^2}\)

\(B=\frac{2a\left|1-2a\right|}{2a-1}\sqrt{5}\)

\(=\frac{2a\left(2a-1\right)}{2a-1}\sqrt{5}=2a\sqrt{5}\)

\(ĐKXĐ:a\ne\frac{1}{2}\)

\(B=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)

\(=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-2a\right)^2}\)

\(=\frac{1}{2a-1}.\sqrt{5}.\sqrt{a^4}.\sqrt{\left(1-2a\right)^2}\)

\(=\frac{1}{2a-1}.\sqrt{5}.a^2.\left|1-2a\right|=\frac{\sqrt{5}.a^2.\left|1-2a\right|}{2a-1}\)

+) Nếu \(a< \frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=1-2a=-\left(2a-1\right)\)

\(\Rightarrow B=\frac{-\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=-\sqrt{5}.a^2\)

+) Nếu \(a>\frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=-\left(1-2a\right)=-1+2a=2a-1\)

\(\Rightarrow B=\frac{\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=\sqrt{5}.a^2\)

\(\frac{2}{2a-1}.\sqrt{5x^4\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}.\sqrt{5x^4\left(2a-1\right)^2}\)

\(=\frac{2}{2a-1}.x^2.\left(2a-1\right).\sqrt{5}\)

\(=2\sqrt{5}x^2\)

Hỏi nhiều thế.

Giải:

\(\dfrac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)

\(=\dfrac{1}{2a-1}.\sqrt{5a^4}.\sqrt{1-4a+4a^2}\)

\(=\dfrac{1}{2a-1}.a^2\sqrt{5}.\sqrt{\left(1-2a\right)^2}\)

\(=\dfrac{1}{2a-1}.a^2\sqrt{5}.\left|1-2a\right|\)

\(=\dfrac{\left|2a-1\right|.a^2\sqrt{5}}{2a-1}\left(1\right)\)

Chắc đề thiếu điều kiện, mình cho thêm để ra kết quả đẹp

ĐK: \(a\ge1\Leftrightarrow2a\ge2\Leftrightarrow2a-1\ge1>0\)

\(\left(1\right)=\dfrac{\left(2a-1\right).a^2\sqrt{5}}{2a-1}\)

\(=a^2\sqrt{5}\)

Vậy ...

Ta có: \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\frac{2}{2a-1}.\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.a\sqrt{5}.\left(2a-1\right)=2a\sqrt{5}\)