\(a.\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}\)

\(b...">

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AH
Akai Haruma
Giáo viên
18 tháng 6 2019

Lời giải:

a)
\((\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}})^2=5-2\sqrt{5}+5+2\sqrt{5}+2\sqrt{(5-2\sqrt{5})(5+2\sqrt{5})}\)

\(=10+2\sqrt{5^2-(2\sqrt{5})^2}=10+2\sqrt{5}\)

\(\Rightarrow \sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}=\sqrt{10+2\sqrt{5}}\)

b)

\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)

\(=\sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}\)

\(=2-\sqrt{3}+2+\sqrt{3}=4\)

c)

\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}\)

\(=\sqrt{12+1-2\sqrt{12}}+\sqrt{12+1+2\sqrt{12}}=\sqrt{(\sqrt{12}-1)^2}+\sqrt{(\sqrt{12}+1)^2}\)

\(=\sqrt{12}-1+\sqrt{12}+1=2\sqrt{12}=4\sqrt{3}\)

AH
Akai Haruma
Giáo viên
26 tháng 6 2019

Lời giải:

a)
\((\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}})^2=5-2\sqrt{5}+5+2\sqrt{5}+2\sqrt{(5-2\sqrt{5})(5+2\sqrt{5})}\)

\(=10+2\sqrt{5^2-(2\sqrt{5})^2}=10+2\sqrt{5}\)

\(\Rightarrow \sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}=\sqrt{10+2\sqrt{5}}\)

b)

\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)

\(=\sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}\)

\(=2-\sqrt{3}+2+\sqrt{3}=4\)

c)

\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}\)

\(=\sqrt{12+1-2\sqrt{12}}+\sqrt{12+1+2\sqrt{12}}=\sqrt{(\sqrt{12}-1)^2}+\sqrt{(\sqrt{12}+1)^2}\)

\(=\sqrt{12}-1+\sqrt{12}+1=2\sqrt{12}=4\sqrt{3}\)

13 tháng 5 2018

a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)

\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left|2\sqrt{3}-1\right|+\left|2-\sqrt{3}\right|\)

\(=2\sqrt{3}-1+2-\sqrt{3}=\sqrt{3}+1\)

b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)

c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

\(=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)

d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)

e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)

f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)

7 tháng 7 2018

\(1.\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\)

\(2.\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}\)

\(3.\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}-\sqrt{3}\)

\(4.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)

\(5.\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{4+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)

\(6.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)

28 tháng 7 2016

a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{20-2\cdot3\cdot\sqrt{20}+9}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{20}+3}}\)

\(=\sqrt{5-\sqrt{6-\sqrt{20}}}\)

\(=\sqrt{5-\sqrt{5-2\sqrt{5}+1}}\)

\(=\sqrt{5-\sqrt{\left(\sqrt{5}+1\right)^2}}\)

\(=\sqrt{5-\sqrt{5}-1}\)

\(=\sqrt{4-\sqrt{5}}\)

c)\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

\(=3-2=1\)

d)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)

\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)

\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)

\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{3}-1+\sqrt{3+1}\)

\(=2\sqrt{3}\)

 

 

 

 

14 tháng 7 2019

\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)

\(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

==================================================

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

===========================================================

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

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9 tháng 9 2016

Bài 2 : 

a,\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12=>\sqrt{24}+\sqrt{45}< 12\)

b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)

c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)

 

7 tháng 7 2017

\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)

các câu còn lại làm tương tự nhé bạn !

19 tháng 8 2017

Hà Nam răng từ\(\sqrt{4}.....\)sang đc 2+ căn 3 đó ???

a: \(=\sqrt{4+2+\sqrt{3}}=\sqrt{6+\sqrt{3}}\)

c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

d: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

\(=\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\)

TH1: x>=2

\(D=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)

TH2: 0<=x<2

\(D=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)