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25 tháng 9 2021
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
27 tháng 7 2021
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
23 tháng 6 2017
= - 0,3288755607 nha 
Hà Phạm Như Ý ! ! !
K VÀ KB NHA ! ! !
NT
1
29 tháng 8 2018
Ta có:
\(\dfrac{\sqrt[4]{17+12\sqrt{2}} +\sqrt[4]{17-12\sqrt{2}}}{2}\)
\(=\dfrac{\sqrt[4]{3^2+2.3.(2\sqrt{2})+\left(2\sqrt{2}\right)^2}+\sqrt[4]{3^2-2.3.(2\sqrt{2})+\left(2\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt[4]{\left(3+2\sqrt{2}\right)^2}+\sqrt[4]{\left(3-2\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt[4]{\left(2+2\sqrt{2}+1\right)^2}+\sqrt[4]{\left(2-2\sqrt{2}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt[4]{[\left(\sqrt{2}+1\right)^2]^2}+\sqrt[4]{[\left(\sqrt{2}-1\right)^2]^2}}{2}\)
\(=\dfrac{\sqrt[4]{\left(\sqrt{2}+1\right)^4}+\sqrt[4]{\left(\sqrt{2}-1\right)^4}}{2}\)
\(=\dfrac{\sqrt{2}+1+\sqrt{2}-1}{2}\)
\(=\dfrac{2\sqrt{2}}{2}\)
\(=\sqrt{2}\) (đpcm)
RK
22 tháng 6 2021
a) A= \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
Vì \(\left\{{}\begin{matrix}2=\sqrt{4}< \sqrt{5}\\2\sqrt{2}=\sqrt{8}>\sqrt{5}\end{matrix}\right.\) nên A = \(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
= \(\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\)
= \(2\left(\sqrt{2}-1\right)\)
RK
22 tháng 6 2021
b) B = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\) (B > 0)
Ta có:
B2 = \(6+2\sqrt{5}-2\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}+6-2\sqrt{5}\)
= \(12-2\sqrt{36-20}\)
= \(12-8\)
= \(4\)
\(\Rightarrow\) B =\(\pm2\) nhưng vì B > 0 nên B = 2
Vậy B = 2
6 tháng 7 2017
a,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{2^2+2\cdot2\cdot\left(2\sqrt{5}\right)+\left(2\sqrt{5}\right)^2}\) \(+\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\sqrt{5}+2^2}=\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)=\(2+2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}\)
b,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=3-2\sqrt{2}+2\sqrt{2}+1=4\)
c,\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
17 tháng 7 2017
a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)
b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)
c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)
d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)
S
22 tháng 6 2019
\(a,\sqrt{15+6\sqrt{6}}=\sqrt{9+6\sqrt{6}+6}=\left(3+\sqrt{6}\right)^2\)
\(b,Dat:\left\{{}\begin{matrix}\sqrt{17-12\sqrt{2}}=a\\\sqrt{17+12\sqrt{2}}=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ab=1\\a^2+b^2=34\end{matrix}\right.\Rightarrow a^2+b^2+2ab=\left(a+b\right)^2=36=\left(\pm6\right)^2.\Rightarrow\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}=6\left(\sqrt{17-12\sqrt{2}};\sqrt{17+12\sqrt{2}}\ge0\right)\)
\(c,=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\left(\sqrt{2}>1=\sqrt{1};\sqrt{3}>1=\sqrt{1};2=\sqrt{4}>\sqrt{3}\right)\)
A = \(\sqrt[4]{3^2+2.3.\left(2\sqrt{2}\right)+\left(2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt[4]{\left(3+2\sqrt{2}\right)^2}-\sqrt{2}\)
A = \(\sqrt{3+2\sqrt{2}}-\sqrt{2}=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}.1+1}-\sqrt{2}=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{2}\)
A = \(\sqrt{2}+1-\sqrt{2}=1\)
\(\sqrt[4]{\left(3+2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{3+2\sqrt{2}}-\sqrt{2}=\sqrt{\left(1+\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)