bài 1:
a) D = \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b) E = \(\sqrt[3]{\sqrt{5}-2}+\sqrt[3]{\sqrt{5}+2}\)
c) F =\(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)
bài 2:
a) C = \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\)
b) D = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\frac{1}{2-\sqrt{3}}\)
c) E =\(\frac{3-x^2}{x+\sqrt{3}}\) với x\(\ne-\sqrt{3}\)
d) F = \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2019}+\sqrt{2020}}\)
e) G = \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}\) (có vô hạn dấu căn)
a/ \(D\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\Rightarrow D=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
b/\(2E=\sqrt[3]{8\sqrt{5}-16}+\sqrt[3]{8\sqrt{5}+16}\)
\(=\sqrt[3]{5\sqrt{5}-3.5.1+3\sqrt{5}-1}+\sqrt[3]{5\sqrt{5}+3.5.1+3\sqrt{5}+1}\)
\(=\sqrt[3]{\left(\sqrt{5}-1\right)^3}+\sqrt[3]{\left(\sqrt{5}+1\right)^3}=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(\Rightarrow E=\sqrt{5}\)
c/
\(F=\sqrt[3]{182+25\sqrt{53}}+\sqrt[3]{182-25\sqrt{53}}\)
\(F^3=364+3F\sqrt[3]{182^2-33125}=364-3F\)
\(\Leftrightarrow F^3+3F-364=0\)
\(\Leftrightarrow\left(F-7\right)\left(F^2+7F+52\right)=0\)
\(\Rightarrow F=7\)
Bài 2:
a/ \(C=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\)
\(=\sqrt{4}-1=2-1=1\)
Bài 2
b/
\(D=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
c/
\(E=\frac{\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)}{x+\sqrt{3}}=\sqrt{3}-x\)
d/
\(F=\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{2020}-\sqrt{2019}}{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)
\(=\sqrt{2020}-1\)
e/
\(G=\sqrt{2+\sqrt{2+\sqrt{2+...}}}\) (G>0)
\(\Rightarrow G^2=2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}\)
\(\Rightarrow G^2=2+G\)
\(\Rightarrow G^2-G-2=0\Rightarrow\left(G+1\right)\left(G-2\right)=0\)
\(\Rightarrow G=2\)