\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt...">
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7 tháng 7 2018

\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\Rightarrow\)\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

                   \(=\frac{2+\sqrt{3}}{2+\left(\sqrt{3}+1\right)}+\frac{2-\sqrt{3}}{2-\left(\sqrt{3}-1\right)}\)

                   \(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)       

                   \(=\frac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

                  \(=\frac{\sqrt{3}+1}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\frac{\sqrt{3}-1}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

                    \(=\frac{2\sqrt{3}}{2\sqrt{3}}=1\)

                          

24 tháng 10 2018

trả lời nhanh hộ t nhé cc :)

24 tháng 10 2018

\(\frac{5\left(\sqrt{6}-1\right)\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}\)

\(=\frac{5\left(\sqrt{6}-1\right)^2}{5}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{1}+\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=\left(\sqrt{6}-1\right)^2-\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{2}-1\right)\)

\(=6-2\sqrt{6}+1-2+2\sqrt{6}-3+\sqrt{2}-1=\sqrt{2}\)

27 tháng 6 2016

\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)\(=\frac{4+2\sqrt{3}}{\sqrt{4}+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{\sqrt{4}-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{4+2\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{4-2\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)\(=\frac{4+2\sqrt{3}}{2+\sqrt{3}+1}+\frac{4-2\sqrt{3}}{2-\sqrt{3}+1}\)

\(=\frac{\left(\sqrt{3}+1\right)^2}{3+\sqrt{3}}+\frac{\left(\sqrt{3}-1\right)^2}{3-\sqrt{3}}\)

\(=\frac{\left(\sqrt{3}+1\right)^2}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\left(\sqrt{3}-1\right)^2}{\sqrt{3}\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}\)

\(=\frac{2\sqrt{3}}{\sqrt{3}}=2\)

3 tháng 7 2017

a,

\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)

=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)

=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)

=\(\sqrt{2}+\sqrt{2}-1\)

=\(2\sqrt{2}-1\)

còn tiếp

3 tháng 7 2017

b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)

=\(6-1+\sqrt{3}-\sqrt{6}\)

=\(5+\sqrt{3}+\sqrt{6}\)