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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) = \(5\sqrt{2}-3\sqrt{6}+3\sqrt{2}+5\sqrt{6}\)
= \(8\sqrt{2}+2\sqrt{6}\)
b) = \(2\sqrt{3}-4\sqrt{2}-5\sqrt{3}-\sqrt{2}\)
= \(-3\sqrt{3}-5\sqrt{2}\)
c) = \(\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=\(\frac{2\sqrt{2}+2-2-\sqrt{2}}{2^2-\sqrt{2^2}}\)
=\(\frac{\sqrt{2}}{4-2}\) = \(\frac{\sqrt{2}}{2}\)
d) = \(2\sqrt{6}-5\sqrt{6}+2\sqrt{2}\)
=\(-3\sqrt{6}+2\sqrt{2}\)
e) = \(8\sqrt{6}+3\sqrt{6}-6\sqrt{6}=5\sqrt{6}\)
f) = \(4\sqrt{3}+9\sqrt{3}-4\sqrt{3}=9\sqrt{3}\)
g) = \(10+5\sqrt{10}-5\sqrt{10}=10\)
h) = \(\frac{\left(3+\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{\left(3-\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
= \(\frac{9+3\sqrt{3}+3\sqrt{3}+3}{3^2-\sqrt{3^2}}+\frac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-\sqrt{3^2}}\)
= \(\frac{12+6\sqrt{3}}{9-3}+\frac{12-6\sqrt{3}}{9-3}\)
= \(\frac{12+6\sqrt{3}+12-6\sqrt{3}}{6}\)
= \(\frac{24}{6}=4\)
k) = \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
l) = \(\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\)
= \(\frac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{\sqrt{8^2}-2^2}\)
= \(\frac{2\sqrt{6}}{8-4}=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\)
a) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(\sqrt{9\cdot11}-\sqrt{9\cdot2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=3\cdot11-3\sqrt{22}-11+3\sqrt{22}\)
\(=33-11=22\)
b)\(3\sqrt{\frac{9}{8}}-\sqrt{\frac{49}{2}}+\sqrt{\frac{25}{18}}\)
\(=\frac{9}{\sqrt{8}}-\frac{7}{\sqrt{2}}+\frac{5}{\sqrt{18}}\)
\(=\frac{9}{2\sqrt{2}}-\frac{7}{\sqrt{2}}+\frac{5}{3\sqrt{2}}\)
\(=\frac{27-42+10}{6\sqrt{2}}\)
\(=-\frac{5}{6\sqrt{2}}\)
c)\(\left(1+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\frac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(=\left(1-\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}+1\right)\)
\(=\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)
\(=1-5=-4\)
Lời giải:
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}}=\sqrt{6+2\sqrt{5-(\sqrt{12}+1)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}\)
\(=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}=\sqrt{(\sqrt{3}+1)^2}\)
\(=\sqrt{3}+1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}(2-\sqrt{3})}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\sqrt{3}\)
\(=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\sqrt{3}=\frac{2(\sqrt{3}+1)}{3-1}+\sqrt{3}=\sqrt{3}+1+\sqrt{3}=2\sqrt{3}+1\)
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{4-\sqrt{12}}}\\ =\sqrt{6+2\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\left(\sqrt{3}-1\right)}\\ =\sqrt{6-2\sqrt{3}-2}\\ =\sqrt{4-2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}\\ =\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\\ =\frac{2}{\sqrt{3}-1}+\sqrt{3}\\ =\frac{2+\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\frac{5-\sqrt{3}}{\sqrt{3}-1}\\ =\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1}\\ -\frac{6\sqrt{3}-8}{2}=3\sqrt{3}-4\)
(bạn nhớ ktr đã nha)