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9 tháng 9 2023

\(2\sqrt[]{37+20\sqrt[]{3}}-\sqrt[]{73-40\sqrt[]{3}}\)

\(=2\sqrt[]{25+2.5.2\sqrt[]{3}+12}-\sqrt[]{48-2.5.4\sqrt[]{3}+25}\)

\(=2\sqrt[]{\left(5+2\sqrt[]{3}\right)^2}-\sqrt[]{\left(5-4\sqrt[]{3}\right)^2}\)

\(=2\left|5+2\sqrt[]{3}\right|-\left|5-4\sqrt[]{3}\right|\)

\(=2\left(5+2\sqrt[]{3}\right)-\left(4\sqrt[]{3}-5\right)\left(vì.4\sqrt[]{3}>5\right)\)

\(=10+4\sqrt[]{3}-4\sqrt[]{3}+5\)

\(=15\)

9 tháng 9 2023

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13 tháng 9 2017

\(\sqrt{37-20\sqrt{3}}+\sqrt{37+20\sqrt{3}}\)

\(=\sqrt{37-2\sqrt{300}}+\sqrt{37+2\sqrt{300}}\)

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

\(=|5-\sqrt{12}|+|5+\sqrt{12}|\)

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

\(=10\)

7 tháng 1 2021

\(A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}-\sqrt{37-20\sqrt{3}}\)

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

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

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

9 tháng 7 2018

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

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

c) \(2\sqrt{5}-3\sqrt{45}+\sqrt{500}=2\sqrt{5}-9\sqrt{5}+10\sqrt{5}=3\sqrt{5}\)

d) \(\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\sqrt{3}=\dfrac{1-\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1-3-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-2-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=-\sqrt{2}\)

e) \(\dfrac{1}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}+5\sqrt{3}=\dfrac{2-\sqrt{3}-\left(2+\sqrt{3}\right)}{4-3}+5\sqrt{3}=-2\sqrt{3}+5\sqrt{3}=3\sqrt{3}\)

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

g) \(\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\sqrt{5}-\dfrac{4}{\sqrt{5}+1}=\dfrac{5+\sqrt{5}-4}{\sqrt{5}+1}=1\)

h)\(\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}=\sqrt{37-20\sqrt{3}+\left(5+2\sqrt{3}\right)}=\sqrt{42-18\sqrt{3}}=\sqrt{\left(3\sqrt{3}+3\right)^2+6}\)

16 tháng 9 2019

\(\sqrt{12-2.2\sqrt{3}.5+25}-\sqrt{12+2.2\sqrt{3}.5+25}\)

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

\(|2\sqrt{3}-5|-2\sqrt{3}-5\)

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

28 tháng 7 2020

bây giờ vẫn còn công chúa

NV
17 tháng 9 2019

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

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

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

\(=10\)

a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)

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

b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)

30 tháng 5 2023

cảm ơn anh ạ