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5 tháng 10 2020
b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
26 tháng 7 2019
a) \(\frac{3}{2\sqrt{3}}=\frac{3}{2.3^{\frac{1}{2}}}=\frac{3^{1-\frac{1}{2}}}{2}=\frac{3^{\frac{1}{2}}}{2}=\frac{\sqrt{3}}{2}\)
b) \(\frac{5}{2\sqrt{3}}=\frac{5\sqrt{3}}{2\sqrt{3}.\sqrt{3}}=\frac{5\sqrt{3}}{6}\)
22 tháng 7 2016
a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1=2\)
=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)
b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
=> B=\(\sqrt{5}+1\)
c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)
=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)
=> A=\(\sqrt{5}\)
Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
= \(A-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1
H
1 tháng 10 2021
\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{15+2.3.\sqrt{6}}\)\(-\sqrt{10+2.2\sqrt{6}}\)
\(=\sqrt{9+2.3\sqrt{6}+6}\)\(-\sqrt{6+2.\sqrt{6}.2+4}\)
\(=\sqrt{\left(3+\sqrt{6}\right)^2}\)\(-\sqrt{\left(\sqrt{6}+2\right)^2}\)
\(=3+\sqrt{6}\)\(-2\)\(-\sqrt{6}=\left(3-2\right)+\left(\sqrt{6}-\sqrt{6}\right)\)
\(=1+0=1\)
2 tháng 10 2021
a) \((\sqrt{3}-\sqrt{2}).\sqrt{(\sqrt{3}+\sqrt{2})^2}\)
\(\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right)\)
\(\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)\(=3-2=1\)
b) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
=\(\sqrt{(2+2\sqrt{5})^2}+\sqrt{(\sqrt{5}-2)^2}\)
=\(2+2\sqrt{5}+\sqrt{5}-2\)\(=3\sqrt{5}\)
2 tháng 10 2018
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{2}}\right)\left(4-\sqrt{10+2\sqrt{2}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5.1}+1}=8+2\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(B=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=-\frac{1}{4}\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+....+\sqrt{2001}-\sqrt{2005}\right)\)
\(=-\frac{1}{4}\left(1-\sqrt{2005}\right)\)
\(=10,94430659\)
\(\text{Lm hơi vắn tắt thông cảm nha!!}\)
A2=\(4+\sqrt{5}+4-\sqrt{5}-2\sqrt{16-5}\)
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