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Ta có: \(\sqrt{2+\sqrt{3}}=\frac{1}{\sqrt{2}}.\sqrt{4+2\sqrt{3}}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}\)
=> \(A=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}}=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}}=\frac{\sqrt{3}+1}{2}\)
\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}.\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}.\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{\sqrt{2}.\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}.\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{\sqrt{2}.\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\)
\(=\sqrt{2}-\frac{\sqrt{2}}{3+\sqrt{3}}+\sqrt{2}-\frac{\sqrt{2}}{3-\sqrt{3}}\)
\(=2\sqrt{2}-\left(\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}\right)\)
\(=2\sqrt{2}-\frac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}\)
\(=2\sqrt{2}-\frac{6\sqrt{2}}{6}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
A=\(\sqrt{2}\), cái kết quả này bấm máy tính là ra được, quan trọng là phải làm thế nào để ra
=\(\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
=\(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{2}}:\left(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{6}}\right)\)
=\(\frac{\sqrt{3}+1}{2\sqrt{2}}:\left(\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}\right)\)
=\(\frac{\sqrt{3}+1}{2\sqrt{2}}:\frac{\sqrt{3}.\left(\sqrt{3}+1\right)-2.2+\sqrt{3}+1}{2\sqrt{6}}\)
=\(\frac{\sqrt{3}+1}{2\sqrt{2}.}.\frac{2\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+1}{2}\)
\(A=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2+\sqrt{3}}\right)}{2-2-\sqrt{3}}+\dfrac{\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)}{2-2+\sqrt{3}}\)
\(=\dfrac{-2\sqrt{2}+\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{6}+\sqrt{6+3\sqrt{3}}}{\sqrt{3}}+\dfrac{2\sqrt{2}+\sqrt{2}\left(\sqrt{3}-1\right)-\sqrt{6}-\sqrt{6-3\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{6+3\sqrt{3}}+\sqrt{6}-\sqrt{2}-\sqrt{6}-\sqrt{6-3\sqrt{3}}}{\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)