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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) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)
= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)
= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
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. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)
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. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\left|\sqrt{3}-1\right|=2-\sqrt{3}+\sqrt{3}-1=1\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
a) \(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2}=\frac{\sqrt{3-2\sqrt{3}+1}}{2}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}\)
\(=\frac{\left|\sqrt{3}-1\right|}{2}=\frac{\sqrt{3}-1}{2}\)
b) \(\sqrt{8}\cdot\sqrt{3-\sqrt{5}}=\sqrt{4}\cdot\sqrt{6-2\sqrt{5}}=2\sqrt{5-2\sqrt{5}+1}=2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\cdot\left|\sqrt{5}-1\right|=2\left(\sqrt{5}-1\right)=2\sqrt{5}-2\)
a)
\((2\sqrt{5}-\sqrt{7})(2\sqrt{5}+\sqrt{7})=(2\sqrt{5})^2-(\sqrt{7})^2=13\)
b)
\((\sqrt{5-2\sqrt{6}}+\sqrt{2})\sqrt{3}=(\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2})\sqrt{3}\)
\(=(\sqrt{(\sqrt{3}-\sqrt{2})^2}+\sqrt{2})\sqrt{3}=(\sqrt{3}-\sqrt{2}+\sqrt{2})\sqrt{3}=\sqrt{3}.\sqrt{3}=3\)
c)
\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)
\(=\sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}=2-\sqrt{3}+2+\sqrt{3}=4\)
d)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{3^2+6-2.3\sqrt{6}}+\sqrt{9+24-2\sqrt{9.24}}\)
\(=\sqrt{(3-\sqrt{6})^2}+\sqrt{(\sqrt{24}-3)^2}=3-\sqrt{6}+\sqrt{24}-3\)
\(=\sqrt{6}\)
e)
\(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}=\sqrt{\frac{6+2\sqrt{5}}{2}}+\sqrt{\frac{6-2\sqrt{5}}{2}}\)
\(=\sqrt{\frac{5+1+2\sqrt{5.1}}{2}}+\sqrt{\frac{5+1-2\sqrt{5.1}}{2}}=\sqrt{\frac{(\sqrt{5}+1)^2}{2}}+\sqrt{\frac{(\sqrt{5}-1)^2}{2}}\)
\(=\frac{\sqrt{5}+1}{\sqrt{2}}+\frac{\sqrt{5}-1}{\sqrt{2}}=\sqrt{10}\)
g)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{20+3-2\sqrt{20.3}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{20}-\sqrt{3})^2}\)
\(=\sqrt{5}-\sqrt{3}-(\sqrt{20}-\sqrt{3})=\sqrt{5}-\sqrt{20}=-\sqrt{5}\)
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=2-\sqrt{3}+\sqrt{3-2\sqrt{3}+1}\left(\text{vì }2>\sqrt{3}\right)\)
\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\sqrt{3}-1\left(\text{vì }\sqrt{3}>1\right)\)
\(=1\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{\left(2\sqrt{6}\right)^2-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3\left(\text{vì }3>\sqrt{6};2\sqrt{6}>3\right)\)
\(=\sqrt{6}\)
a/
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
\(=1\)
b/
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}\)
\(=3-\sqrt{6}+\sqrt{24}-3\)
\(=-\sqrt{6}+2\sqrt{6}=\sqrt{6}\)
tick cho mình nha