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3 tháng 8 2019
\(\sqrt{3}+\sqrt{8-2\sqrt{15}}\\ =\sqrt{3}+\sqrt{5-2\sqrt{5\cdot3}+3}\\ =\sqrt{3}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{5}\)
\(\sqrt{x-1-2\sqrt{x-2}}\left(x\ge2\right)\\ =\sqrt{x-2-2\sqrt{x-2}+1}\\ =\sqrt{\left(\sqrt{x-2}-1\right)^2}\\ =\left|\sqrt{x-2}-1\right|\\ =\left[{}\begin{matrix}\sqrt{x-2}-1\left(\sqrt{x-2}\ge1\Leftrightarrow x\ge3\right)\\1-\sqrt{x-2}\left(\sqrt{x-2}< 1\Leftrightarrow2\le x< 3\right)\end{matrix}\right.\)
Chúc bạn học tốt nha
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12 tháng 12 2018
a,\(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(=3-1\)
\(=2\)
b, \(\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3}+\sqrt{5}-3\sqrt{3}+\sqrt{5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{2\sqrt{5}-2\sqrt{3}}{\sqrt{2}}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(5-3\right)\)
\(=2\sqrt{2}\)
28 tháng 10 2018
\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{\sqrt{9}.\sqrt{2}-\sqrt{4}.\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{(\sqrt{3}-\sqrt{2})\sqrt{6^2}}{\sqrt{3}-\sqrt{2}}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\sqrt{6}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{6+6\sqrt{6}-5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{1+6\sqrt{6}}{1+\sqrt{6}}\)
11 tháng 8 2015
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{6-2+2\sqrt{3}}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(1+\sqrt{3}\right)^2}=1+\sqrt{3}\)
ON
27 tháng 8 2018
Bạn để ý nhé cách tính là nhân cả tử và mẫu với căn 2
1) \(=\frac{\sqrt{2}.\sqrt{5-\sqrt{21}}}{\sqrt{2}}\)
\(=\frac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\sqrt{21}+3}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)
\(=\frac{\sqrt{14}-\sqrt{6}}{2}\)
câu 2 bạn làm tương tự nhé
\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\frac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{2.\left(2-\sqrt{3}\right)}}{\sqrt{2}}+\frac{\sqrt{2.\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}}\)
\(=\frac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}+\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}+\frac{\left|\sqrt{3}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}.\sqrt{2}}{\left(\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{6}}{2}\)
\(=\sqrt{6}\)
Chúc bạn hok tốt!!! The Godlin