\(\left(3a+4\right)^2+\left(4a-1\right)^2+\left(2+5a\right)\left(2-5a\right)=9a^2+24a+16+16a^2-8a+1+4-25a^2=16a+21\)

a: \(Q=\left(\dfrac{a^2+4a+4-a^2+4a-4+4a^2}{\left(a-2\right)\left(a+2\right)}\right):\dfrac{a\left(a-3\right)}{5a\left(2-a\right)}\)

\(=\dfrac{4a^2+8a}{\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-5\left(a-2\right)}{a-3}\)

\(=\dfrac{-20a}{a-3}\)

b: Q chia hết cho 20 thì a/a-3 là số nguyên

=>\(a-3\in\left\{1;-1;3;-3\right\}\)

=>a=4 hoặc a=6

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

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

Xét \(a>\dfrac{1}{2}\Rightarrow1-2a< 0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

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

Xét \(a< \dfrac{1}{2}\Rightarrow1-2a>0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

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

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

- Với \(2a-1>0\Rightarrow a>\dfrac{1}{2}\) thì \(E=\dfrac{a^2\left(2a-1\right).\sqrt{5}}{2a-1}=a^2\sqrt{5}\)

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

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

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

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

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

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

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

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

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

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

a)\(\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

b)\(\sqrt{\left(2-\sqrt{11}\right)^2}=2-\sqrt{11}\)

c)\(2\sqrt{a^2}=2a\) vì a≥0