rút gọn
A= \(\frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}-\sqrt{5}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
B= \(\frac{5+2\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
K
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20 tháng 5 2016
\(2B=5\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{5}\right)^2+\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}-\sqrt{5}\right)^2 \)
\(=5\left(\left(\sqrt{3}+1\right)+\left(\sqrt{5}-1\right)-\sqrt{5}\right)^2+\left(\left(\sqrt{3}-1\right)+\left(\sqrt{5}+1\right)-\sqrt{5}\right)^2\)
\(=5\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^2=5.3+3=18\)
\(\Rightarrow B=9\)
1 tháng 3 2017
Xem kỹ lại đề nhé! loại này đề lệch một tý thôi -->Không rút được !
p/s: Tránh truongf hợp làm đến cuối mới biết đề sai.
12 tháng 8 2020
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
\(A=\frac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}+\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\\ A=\frac{3+2\sqrt{15}+5}{3-5}+\frac{5-2\sqrt{15}+3}{5-3}\\ A=\frac{3+2\sqrt{15}+5-\left(5-2\sqrt{15}+3\right)}{-2}\\ A=\frac{4\sqrt{15}}{-2}=-2\sqrt{15}\)
\(B=\frac{\sqrt{5}\left(5+2\sqrt{5}\right)}{\left(\sqrt{5}\right)^2}+\frac{\sqrt{3}\left(3+\sqrt{3}\right)}{\left(\sqrt{3}\right)^2}-\left(\sqrt{5}+\sqrt{3}\right)\\ B=\frac{5\left(\sqrt{5}+2\right)}{5}+\frac{3\left(\sqrt{3}+1\right)}{3}-\left(\sqrt{5}+\sqrt{3}\right)\\ B=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}\\ B=3\)