Rút gọn:
\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
K
Khách
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2 tháng 10 2021
\(A=\dfrac{\sqrt{6+2\sqrt{5}}}{2-\sqrt{6-2\sqrt{5}}}-\dfrac{\sqrt{6-2\sqrt{5}}}{2+\sqrt{6+2\sqrt{5}}}\)
\(=\dfrac{\sqrt{5}+1}{2-\sqrt{5}+1}-\dfrac{\sqrt{5}-1}{3+\sqrt{5}}\)
\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)
\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)
\(=4\)
4 tháng 7 2023
\(=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{5}+1}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{5}+1}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}\)
\(=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{4}\)
\(=\sqrt{2}\cdot\dfrac{2}{4}=\dfrac{1}{\sqrt{2}}\)
6 tháng 7 2021
a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow A^2=2-\sqrt{3}+2+\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=4-2\sqrt{4-3}=4-2=2\)
\(\Rightarrow A=-\sqrt{2}\)
b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)
\(=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow B=\sqrt{2}\)
HT
\(\frac{\sqrt{5}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{2\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)Rút gọn
1
26 tháng 12 2016
=\(\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
=\(\frac{5-\sqrt{15}-2\sqrt{15}+6-10-2\sqrt{15}-\sqrt{15}-3}{5-3}\)
\(=\frac{-2-6\sqrt{15}}{2}=\frac{-2\left(1+3\sqrt{15}\right)}{3}=-1-3\sqrt{15}\)
AH
Akai Haruma
Giáo viên
11 tháng 8 2021
Lời giải:
1/
\(=\frac{3.\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)
2/
\(=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}=1\)
3/
\(=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)
Y
30 tháng 5 2023
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)
27 tháng 2 2018
\(A=\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{6-2\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{5+2\sqrt{5}+1}}+\frac{6-2\sqrt{5}}{2+\sqrt{5-2\sqrt{5}+1}}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6-2\sqrt{5}}{2+\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\frac{6+2\sqrt{5}}{2+\left|\sqrt{5}+1\right|}+\frac{6-2\sqrt{5}}{2-\left|\sqrt{5}-1\right|}\)
\(=\frac{6+2\sqrt{5}}{2+\sqrt{5}+1}+\frac{6-2\sqrt{5}}{2-\sqrt{5}+1}\)( vì \(\sqrt{5}+1>0;\sqrt{5}-1>0\))
\(=\frac{6+2\sqrt{5}}{3+\sqrt{5}}+\frac{6-2\sqrt{5}}{3-\sqrt{5}}\)
\(=2+2\)
\(=4\)
Vậy A = 4
Tích cho mk nhoa !!!! ~~
25 tháng 8 2021
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
Ta có: \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}\)
=0