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a) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{\left(3-\sqrt{2}\right)+\left(3+\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-\left(\sqrt{2}\right)^2}=\frac{6}{7}\)
b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}=\frac{2\left(2\sqrt{3}+3\sqrt{3}\right)-3\left(3\sqrt{2}-3\sqrt{3}\right)}{\left(3\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{3}\right)}=\frac{19\sqrt{3}-9\sqrt{2}}{-45+15\sqrt{6}}=-\frac{13\sqrt{3}+10\sqrt{2}}{15}\)c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{2}=\frac{16}{2}=8\)d) \(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}=\frac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=-\frac{18\sqrt{3}}{19}\)
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
h)
\(H=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2-\sqrt{3})(2+\sqrt{3})}}=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=2\sqrt{3}\)
i)
\(I=\frac{2+\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3.1}}}+\frac{2-\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3.1}}}=\frac{2+\sqrt{3}}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{(\sqrt{3}-1)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6}{6}=1\)
ê)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+1+2(\sqrt{2}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+1+2(\sqrt{2}+\sqrt{5})}=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}=\sqrt{2}+\sqrt{5}+1\)
g)
\(13+\sqrt{48}=13+2\sqrt{12}=12+1+2\sqrt{12.1}=(\sqrt{12}+1)^2\)
\(\Rightarrow \sqrt{13+\sqrt{48}}=\sqrt{12}+1\)
\(\Rightarrow \sqrt{3+\sqrt{13+\sqrt{48}}}=\sqrt{4+\sqrt{12}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
\(\Rightarrow 2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}=2\sqrt{2-\sqrt{3}}=\sqrt{2}.\sqrt{4-2\sqrt{3}}=\sqrt{2}.\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{2}(\sqrt{3}-1)=\sqrt{6}-\sqrt{2}\)
\(\Rightarrow G=1\)
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\) =\(\frac{6}{6}=1\)
\(\Rightarrow A=\sqrt{2}\)