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22 tháng 7 2017
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
NV
16 tháng 7 2016
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/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
NC
16 tháng 8 2020
1) \(b=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right).\sqrt{2}\)
\(b=2-\sqrt{6-2\sqrt{5}}\)
\(b=2-\sqrt{5-2\sqrt{5}+1}\)
\(b=2-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(b=2-\sqrt{5}+1=3-\sqrt{5}\)
WR
28 tháng 6 2019
=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}.\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{4-3}=1\)
WR
28 tháng 6 2019
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}.\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
TN
14 tháng 8 2018
\(A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(B=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\)
\(C=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(7-5\right)=2\sqrt{9}\)
\(D=\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
\(E=\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{5^2-\sqrt{5}^2}=\dfrac{60}{20}=3\)
3 tháng 11 2019
a) \(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2\sqrt{3}+2-\sqrt{3}\)
\(=\left(2\sqrt{3}-\sqrt{3}\right)+2\)
\(=\sqrt{3}+2\)
b) \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{1+\sqrt{5}}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{1+\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{12}{4}=3\)
c) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{1}=14\)
\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-2\)
\(=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}-\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}-2\)
\(=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{5}-1\right)}^2}{\sqrt{2}}-2\)
\(=\frac{\sqrt{5}+1}{\sqrt{2}}-\frac{\sqrt{5}-1}{\sqrt{2}}-2\)
\(=\frac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}-2\)
\(=\frac{2}{\sqrt{2}}-2\)
\(=\frac{2-2\sqrt{2}}{\sqrt{2}}\)
\(=\frac{\sqrt{2}.\left(\sqrt{2}-2\right)}{\sqrt{2}}\)
\(=\sqrt{2}-2\)
.... Đúng thì ủng hộ nha ....
Đặt A= \(\sqrt{3+\sqrt{5}}\)- \(\sqrt{3-\sqrt{5}}\)- 2
<=>(A+2)^2 = 3+\(\sqrt{5}\)+ 3 - \(\sqrt{5}\)- 2. \(\sqrt{9-5}\)(A+2>0 do \(\sqrt{3+\sqrt{5}}\)> \(\sqrt{3-\sqrt{5}}\))
= 6 - 4 = 2
<=> A+2 = \(\sqrt{2}\) ( vì A+ 2>0)
<=> A= \(\sqrt{2}\)- 2