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a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
\(Q=\left(\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(5-2\right)=-3\)
\(S=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\sqrt{\frac{6+2\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\frac{\sqrt{10}-\sqrt{2}}{4}-\sqrt{\frac{\left(\sqrt{5}+1\right)^2}{4}}\)
\(=\frac{\sqrt{10}-\sqrt{2}}{4}-\frac{\sqrt{5}+1}{2}=\) bạn coi lại đề
\(T=\frac{4\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=\frac{4\left(1+\sqrt{3}\right)}{-2}-\sqrt{3}=-2-3\sqrt{3}\)
đúng là cậu S sai đề thật ;-;;
S = \(\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}\)
ụa, không có ai ạ?
em đang cần gấp lắm ;-;;