Bài 1:
a: \(C=\dfrac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)
\(=\dfrac{\sqrt{x+2+2\cdot\sqrt{x+2}\cdot\sqrt{x-2}+x-2}}{\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}}{\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)}=\dfrac{1}{\sqrt{x+2}}\)
b: Thay \(x=3+2\sqrt{6}\) vào C, ta được:
\(C=\dfrac{1}{\sqrt{3+2\sqrt{6}+2}}=\dfrac{1}{\left(\sqrt{3}+\sqrt{2}\right)}=\sqrt{3}-\sqrt{2}\)
Bài 2:
a: \(C=\dfrac{a}{a-16}-\dfrac{2}{\sqrt{a}-4}-\dfrac{2}{\sqrt{a}+4}\)
\(=\dfrac{a}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\dfrac{2\left(\sqrt{a}+4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\dfrac{2\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\dfrac{a-2\sqrt{a}-8-2\sqrt{a}+8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\dfrac{a-4\sqrt{a}}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\dfrac{\sqrt{a}}{\sqrt{a}+4}\)
b: Thay \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\) vào C, ta được:
\(C=\dfrac{\sqrt{\left(\sqrt{5}-2\right)^2}}{\sqrt{\left(\sqrt{5}-2\right)^2}+4}=\dfrac{\sqrt{5}-2}{\sqrt{5}-2+4}\)
\(=\dfrac{\sqrt{5}-2}{\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}=\dfrac{9-4\sqrt{5}}{5-4}=9-4\sqrt{5}\)
