a) Ta có: Vì \(x \ge 0\) và \( y\ge 0\) nên \(x+y \ge 0 \Leftrightarrow |x+y|=x+y\).
\(\dfrac{2}{x^2 - y^2}\sqrt {\dfrac{3 (x + y)^2}{2}} =\dfrac{2}{x^2 - y^2}\sqrt {\dfrac{3}{2}.(x+y)^2} \)
\(=\dfrac{2}{x^2 - y^2}.\sqrt{\dfrac{3}{2}}.\sqrt{(x+y)^2}\)
\(=\dfrac{2}{x^2 - y^2}.\sqrt{\dfrac{3}{2}}.|x+y|\)
\(=\dfrac{2}{(x+y)(x-y)}.\sqrt{\dfrac{3}{2}}.(x+y)\)
\(=\dfrac{2}{x-y}.\sqrt{\dfrac{3}{2}}\)
\(=\dfrac{1}{x-y}.2.\sqrt{\dfrac{3}{2}}\)
\(=\dfrac{1}{x-y}.\sqrt{\dfrac{2^2.3}{2}}\)
\(=\dfrac{1}{x-y}.\sqrt{6}\) \(=\dfrac{\sqrt 6}{x-y}\)
b) 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:
\(\dfrac{2}{2a-1}\sqrt{5}.|a|.|1-2a|=\dfrac{2}{2a-1}\sqrt{5}.a.(2a-1)\)\(=2\sqrt{5}a\).
Vậy \(\dfrac{2}{2a-1}\sqrt{5a^2(1-4a+4a^2)}=2\sqrt{5}a\).
