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29 tháng 6 2017

\(\sqrt{15+6\sqrt{6}}-\sqrt{33-12\sqrt{6}}\)

\(\Leftrightarrow\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(\Leftrightarrow3+\sqrt{6}-2\sqrt{6}+3\)

\(\Leftrightarrow6-\sqrt{6}\)

8 tháng 7 2021

Bài này chắc rút gọn

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{3^2+2.3\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)

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

\(=\sqrt{6}\)

23 tháng 8 2021

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{9+6-2.3.\sqrt{6}}+\sqrt{9+24-2.3.2\sqrt{6}}\)

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

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

23 tháng 8 2021

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-2.3\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)

\(=\sqrt{15-2\sqrt{54}}+\sqrt{33-2\sqrt{216}}\)

\(=\sqrt{9-2\sqrt{6}\sqrt{9}+6}+\sqrt{24-2\sqrt{24}\sqrt{9}+\sqrt{9}}\)

\(=3-\sqrt{6}+\sqrt{24}-3\)

\(=\sqrt{24}-\sqrt{6}\)

\(=\sqrt{6}\)

 

9 tháng 10 2021

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)

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

l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)

j: \(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)

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

30 tháng 9 2023

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

\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

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

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

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

\(=1\)

b) \(\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)

\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)

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

\(=\left|2\sqrt{6}-3\right|-\left|5-2\sqrt{6}\right|\)

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

\(=4\sqrt{6}-8\)

c) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)

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

\(=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|\)

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

\(=2\)

30 tháng 9 2023

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

\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)

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

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

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

\(=1\)

\(---\)

\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)

\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)

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

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

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

\(=4\sqrt{6}-8\)

\(---\)

\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)

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

\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|\)

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

\(=2\)

#\(Toru\)

10 tháng 9 2023

yêu cầu ?

10 tháng 9 2023

= 2 căn 6 - 3 + 3 - căn 6

= căn 6

Cho mình xin 1 tick nha trang

22 tháng 5 2017

tui hướng dẫn thui nha,,,,\(\sqrt{33-12\sqrt{6}}=\sqrt{24-2.2\sqrt{6}.3+9}=\sqrt{\left(2\sqrt{6}-3\right)^2}=2\sqrt{6}-3\)

ấy dễ ko???,,,bạn lm tương tự típ nhá,,,,

6 tháng 8 2018

ta có : \(A=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

6 tháng 8 2018

????????????????????

19 tháng 8 2021

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a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

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

=-3

b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)

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

=1

c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

\(=\sqrt{6}\)