A \(=\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
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Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
Đọc tiếp
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Rút Gọn Afrac{left(frac{1}{4}-frac{sqrt{2}}{7}+frac{3sqrt{2}}{35}right)frac{1}{25}}{left(frac{1}{10}+frac{3sqrt{2}}{35}-frac{sqrt{2}}{5}right)frac{5}{7}} Bfrac{3+sqrt{5}}{sqrt{10}+sqrt{3+sqrt{5}}}-frac{3-sqrt{5}}{sqrt{10}+sqrt{3+sqrt{5}}} Cfrac{3+sqrt{5}}{sqrt{2}+sqrt{3+sqrt{5}}}-frac{3-sqrt{5}}{sqrt{2}-sqrt{3-sqrt{5}}}
Đọc tiếp
Rút Gọn A=\(\frac{\left(\frac{1}{4}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\frac{1}{25}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{35}-\frac{\sqrt{2}}{5}\right)\frac{5}{7}}\)
B=\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
C=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
1
a. frac{10+2sqrt{10}}{sqrt{5}+sqrt{2}}+frac{8}{1-sqrt{5}} b.frac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}-frac{sqrt{5}+sqrt{27}}{sqrt{30}+sqrt{162}} c. sqrt{frac{2-sqrt{3}}{2+sqrt{3}}}+sqrt{frac{2+sqrt{3}}{2-sqrt{3}}}
d. frac{sqrt{3-sqrt{5}}.left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}} e. frac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{1}{sqrt{2}-sqrt{2-sqrt{3}}} f. frac{left(sqrt{5}+2right)^2-8sqrt{5}}{2sqrt{5}-4}
Đọc tiếp
1
a. \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\) b.\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\) c. \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
d. \(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\) e. \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\) f. \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
a/ \(\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
b/ \(\frac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)
c/ \(\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}+\frac{\sqrt{\left(2+\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=2-\sqrt{3}+2+\sqrt{3}=4\)
d/ \(\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{8}=\frac{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}=1\)
e/ \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\frac{\sqrt{2}\left(3-\sqrt{3}+3+\sqrt{3}\right)}{6}=\sqrt{2}\)
f/ \(\frac{9+4\sqrt{5}-8\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)
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\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
Lời giải:
Đặt \(\sqrt{3+\sqrt{5}}=a; \sqrt{3-\sqrt{5}}=b\) và biểu thức đã cho là $P$
\((a+b)^2=6+2\sqrt{(3-\sqrt{5})(3+\sqrt{5})}=10\Rightarrow a+b=\sqrt{10}\)
\((a-b)^2=6-2\sqrt{(3-\sqrt{5})(3+\sqrt{5})}=2\Rightarrow a-b=\sqrt{2}\)
$ab=\sqrt{(3-\sqrt{5})(3+\sqrt{5})}=2$
\(P=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}=\frac{\sqrt{10}(a^2-b^2)+ab(a-b)}{10+\sqrt{10}(a+b)+ab}=\frac{\sqrt{10}.\sqrt{10}.\sqrt{2}+2\sqrt{2}}{10+\sqrt{10}.\sqrt{10}+2}\)
\(=\frac{6\sqrt{2}}{11}\)
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Tính giá trị biểu thức
\(A=\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3}+\sqrt{5}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\\ \\ \)
tinh
a. sqrt{5}-sqrt{48}+5sqrt{27}-sqrt{45}
b.left(sqrt{5}+sqrt{2}right)left(3sqrt{2}-1right)
c.3sqrt{50}-2sqrt{75}-4frac{sqrt{54}}{sqrt{3}}-3sqrt{frac{1}{3}}
d.sqrt{left(sqrt{3}-3right)^2}+sqrt{4-2sqrt{3}}
e.frac{5sqrt{2}-2sqrt{5}}{sqrt{5}-sqrt{2}}+frac{6}{2-sqrt{10}}-frac{20}{sqrt{10}}
f.frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}+frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}+1}{sqrt{5}-1}
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tinh
a. \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)
b.\(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)
c.\(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)
d.\(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)
e.\(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)
f.\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
Tính \(S=\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
Tính giá trị biểu thức:
A=\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3}+\sqrt{5}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3}-\sqrt{5}}\\ \)
Tính :
a, \(B=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
b, \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, \(C=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}-\frac{\sqrt{5-2\sqrt{6}}}{3}\)
\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=8\)
\(A^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)
\(A^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(A^2=\left(\sqrt{5}+1\right)^2\Rightarrow A=\sqrt{5}+1\) (do \(A>0\))
\(C=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}-\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{3}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}-\frac{\sqrt{3}-\sqrt{2}}{3}\)
\(=\frac{\sqrt{2}}{6}+\frac{\sqrt{2}}{3}=\frac{\sqrt{2}}{2}\)
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