Question

Rút gọn:

C= \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+3\sqrt[]{\dfrac{1}{3}}\right)\)
D=\(\left(\dfrac{1}{\sqrt{5}-2}-\dfrac{1}{\sqrt{5+2}}+1\right).\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)

Answer 1

\(C=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+3\sqrt{\dfrac{1}{3}}\right)=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(2+\sqrt{\dfrac{9}{3}}\right)=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}=\sqrt{2}\)

Answer 2

\(D=\left(\dfrac{1}{\sqrt{5}-2}-\dfrac{1}{\sqrt{5}+2}+1\right).\dfrac{1}{\left(\sqrt{2}+1\right)^2}=\left(\dfrac{\sqrt{5}+2-\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+1\right).\dfrac{1}{\left(\sqrt{2}+1\right)^2}=\dfrac{4+1}{5-4}.\dfrac{1}{3+2\sqrt{2}}=\dfrac{5}{3+2\sqrt{2}}=\dfrac{5\left(3-2\sqrt{2}\right)}{9-8}=15-10\sqrt{2}\)