\(\sqrt{ }\)) × (\(\sqrt{\sqrt{ }}\)

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14 tháng 6 2018

Mình làm 5 bài trắc nha

Hỏi đáp Toán

14 tháng 6 2018

Hỏi đáp Toán

10 tháng 6 2019

bấm máy tính là ra hết bn akhaha

10 tháng 6 2019

giải giúp mình đi bạn

26 tháng 7 2018

\(A=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}=\sqrt{19-2\sqrt{90}}-\sqrt{19+2\sqrt{90}}=\sqrt{10-2.\sqrt{10}.3+9}-\sqrt{10+2.\sqrt{10}.3+9}=\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+3\right)^2}=\sqrt{10}-3-\sqrt{10}-3=-6\)\(B=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}+\sqrt{6+2.\sqrt{3}.\sqrt{6}+3}-\sqrt{24+12\sqrt{3}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{\sqrt{3}}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}=\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}=0\)

Y
4 tháng 7 2019

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

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

\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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

\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)

\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)

\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)

\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)

\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)

\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)

\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)

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

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

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

28 tháng 6 2019

a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))

=\(\sqrt{2006}^2-\sqrt{2005}^2\)

=2006-2005

=1

16 tháng 5 2019

câu b:

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

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

\(=4\sqrt{6}-2\sqrt{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}\)

\(=4\sqrt{6}-2\sqrt{5^2-\left(2\sqrt{6}\right)^2}\)

\(=4\sqrt{6}-2\sqrt{25-24}=4\sqrt{6}-2\)

mấy câu khác tương tự

1 tháng 12 2017

1) \(\sqrt{36+12\sqrt{5}}=\sqrt{\left(\sqrt{30}+\sqrt{6}\right)^2}=\sqrt{30}+\sqrt{6}\)

2)\(\sqrt{21-6\sqrt{6}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}=\sqrt{18}-\sqrt{3}\)

3)\(\sqrt{6-2\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{9}-1\right)^2}\)

\(=\sqrt{5}-1-\left(\sqrt{9}-1\right)\)

\(=\sqrt{5}-\sqrt{9}\)

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

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

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

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

6)\(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

7)\(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}=\sqrt{\left(\sqrt{20}-1\right)^2}+\sqrt{\left(\sqrt{20}+1\right)^2}\)

\(=\sqrt{20}-1+\sqrt{20+1}=2\sqrt{20}\)

17 tháng 6 2018

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