Câu 1:
a) \(A=2022+\sqrt{9}-\sqrt{4}=2022+3-2=2023\)
b) \(B=\sqrt{12}+3\sqrt{27}-2\sqrt{75}\) \(=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}\) \(=\sqrt{3}\)
c) \(C=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{3}\) \(=\left|2+\sqrt{3}\right|-\sqrt{3}\) \(=2+\sqrt{3}-\sqrt{3}\) \(=2\)
d) \(D=5\sqrt{9}-\sqrt{36}=15-6=9\)
e) \(G=5\sqrt{7}+\sqrt{28}-\sqrt{63}\) \(=5\sqrt{7}+2\sqrt{7}-3\sqrt{7}=4\sqrt{7}\)
f) \(F=\sqrt{24}+2\sqrt{54}\) \(=2\sqrt{6}+6\sqrt{6}=8\sqrt{6}\)
Câu 2:
a) \(\dfrac{1}{\sqrt{5}+2}=\dfrac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\) \(=\dfrac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\)
b) \(\dfrac{5}{\sqrt{6}-1}=\dfrac{5\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}\) \(=\dfrac{5\left(\sqrt{6}+1\right)}{6-1}\) \(=\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}+1\)
c) \(\dfrac{1}{3+2\sqrt{2}}=\dfrac{3-2\sqrt{2}}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\) \(=\dfrac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}\)
d) \(\dfrac{1}{2-\sqrt{3}}=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\) \(=\dfrac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\)
