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TK
2
12 tháng 4 2017
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)
\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left(3-\sqrt{6}\right)-\left(2\sqrt{6}-3\right)\)
\(=3-\sqrt{6}-2\sqrt{6}+3\)
\(=6-3\sqrt{6}\)
24 tháng 7 2019
Ko vt lại đề nha bn:
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)
\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left(3-\sqrt{6}\right)-\left(2\sqrt{6}-3\right)\)
\(=3-\sqrt{6}-2\sqrt{6}+3\)
\(=6-3\sqrt{6}\)
Rất vui vì giúp đc bn !!!
25 tháng 6 2017
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
= \(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(2-\sqrt{3}+\sqrt{3}-1\) = \(1\)
b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
= \(3-\sqrt{6}+2\sqrt{6}-3\) = \(\sqrt{6}\)
c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)
= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)
= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)
DS
1
2 tháng 4 2015
\(\sqrt{3^2-2.3.\sqrt{6}+6}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2.\sqrt{6}\right)^2}\)
= \(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2.\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|+\left|3-2\sqrt{6}\right|=3-\sqrt{6}-3+2\sqrt{6}=\sqrt{6}\)
4 tháng 9 2016
1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)
4 tháng 9 2016
1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{2}-\left(\sqrt{5}-\sqrt{2}\right)=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)
\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)
14 tháng 8 2019
\(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=1-\sqrt{3}-\sqrt{3}-2\)
\(=-2\sqrt{3}-1\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+4-2\sqrt{3}\)
\(=6-3\sqrt{3}\)
DK
2 tháng 8 2016
*****~~~~~~~~~~*****
\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+\sqrt{6}}{\sqrt{6}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}+1}\)
\(=\sqrt{3}+\sqrt{6}\)
\(=\sqrt{3}\left(1+\sqrt{2}\right)\)
*****~~~~~~~~~~*****
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(=\sqrt{3}+2+\sqrt{2}\)
(Chúc bạn học tốt nha!)
KV
10 tháng 3 2019
\(A=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\) \(\left(\sqrt{6}+11\right)\)
\(A=\left(\frac{3.\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}\right)+\left(\frac{2.\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}{\sqrt{6}-2}\right)-\frac{4.\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\)\(\left(\sqrt{6}+11\right)\)
\(A=\left(3.\left(\sqrt{6}-1\right)+2.\left(\sqrt{6}+2\right)-4.\left(3+\sqrt{6}\right)\right)\left(\sqrt{6}+11\right)\)
\(A=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=-115\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)