Rút gọn :
\(A=\frac{3-4\sqrt{3}}{2-\sqrt{3}}\)
K
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6 tháng 7 2019
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
6 tháng 7 2019
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
KT
12 tháng 8 2018
\(A=\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\frac{6}{\sqrt{3}}\)
\(=6-3\sqrt{3}+4+\sqrt{3}+\frac{6}{\sqrt{3}}\)
\(=10-2\sqrt{3}+\frac{6}{\sqrt{3}}\)
\(=\frac{10\sqrt{3}-6+6\sqrt{3}}{\sqrt{3}}\)
\(=\frac{16\sqrt{3}-6}{\sqrt{3}}\)
NN
6 tháng 9 2020
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.........+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{2-1}{\sqrt{1}+\sqrt{2}}+\frac{3-2}{\sqrt{2}+\sqrt{3}}+........+\frac{2018-2017}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}+......+\)
\(\frac{\left(\sqrt{2018}-\sqrt{2017}\right)\left(\sqrt{2018}+\sqrt{2017}\right)}{\sqrt{2017}+\sqrt{2018}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+........+\left(\sqrt{2018}-\sqrt{2017}\right)\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+......+\sqrt{2018}-\sqrt{2017}\)
\(=-\sqrt{1}+\sqrt{2018}=\sqrt{2018}-\sqrt{1}\)
A=\(\frac{\left(2-\sqrt{3}\right)^2}{2-\sqrt{3}}=\)2-\(\sqrt{3}\)
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