Hãy rút gọn: \(\dfrac{2}{2a - 1}\sqrt {5a^2(1 - 4a + 4a^2} )\) với \(a > 0,5.\)

* Hướng dẫn giải

Ta có:

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

\(=\dfrac{2}{2a-1}\sqrt{5a^2(1-2.2a+2^2a^2)}\)

\(=\dfrac{2}{2a-1}\sqrt{5a^2 [1^2-2.1.2a+(2a)^2]}\)

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

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

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

Vì \(a> 0,5\) nên \(a>0 \Leftrightarrow |a| =a\).

Vì \(a> 0,5 \Leftrightarrow 2a> 2.0,5 \Leftrightarrow 2a >1 \) hay \( 1<2a\)

\(\Leftrightarrow 1-2a < 0 \Leftrightarrow |1-2a|=-(1-2a)\)

\(=-1+2a=2a-1\)

Thay vào trên, ta được: 

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

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