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a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)
\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)
\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)
b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)
\(=4\sqrt{3}-\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)
\(=-\frac{4}{\sqrt{3}}-\frac{10}{\sqrt{3}}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)
c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(=27+12\sqrt{5}+12\sqrt{5}\)
\(=27+24\sqrt{5}\)
d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{6}+2-\sqrt{3}-\sqrt{2}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
\(=4+2\sqrt{3}-2\sqrt{3}+4\)
= 8
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
= 14
a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(=2\sqrt{2}.\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)
= 9 (đpcm)
b) \(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
\(=\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2^{\frac{1}{2}}\left(\sqrt{2}-1\right)}\)
\(=\sqrt{2\left(\sqrt{2}-1\right)}\) (đpcm)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
a.
\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}\)
\(=\frac{\sqrt{3}-\sqrt{1}}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{9}-\sqrt{7}}{9-7}\)
\(=\frac{\sqrt{9}-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{1}}{2}\)
\(=\frac{3-1}{2}=1\)
b.
\(B=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)
c.
\(C=\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{6-1}+\frac{4\sqrt{6}+8}{6-4}-\frac{36+12\sqrt{6}}{9-6}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}-\frac{36+12\sqrt{6}}{3}-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
d)D=\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)( \(x\ge2\))
=\(\sqrt{x+2\sqrt{2}.\sqrt{x-2}}+\sqrt{x-2\sqrt{2}.\sqrt{x-2}}\)
=\(\sqrt{\left(x-2\right)+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}.\sqrt{x-2}+2}\)
=\(\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
=\(\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)(1)
TH1: \(2\le x\le4\)
Từ (1)<=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}\)
=\(2\sqrt{2}\)
TH2. x\(>4\)
Từ (1) <=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{2}+\sqrt{x-2}\)=\(2\sqrt{x-2}\)
Vậy \(\left[{}\begin{matrix}2\le x\le4\\x>4\end{matrix}\right.< =>\left[{}\begin{matrix}D=2\sqrt{2}\\D=2\sqrt{x-2}\end{matrix}\right.\)