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\(A=\frac{\sqrt{3}-1}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{\sqrt{3}-1}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}\)
\(=\frac{\sqrt{3}-1}{1+\frac{\sqrt{4+2\sqrt{3}}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{4-2\sqrt{3}}}{2}}=\frac{\sqrt{3}-1}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}\)
\(=\frac{\sqrt{3}-1}{\frac{3+\sqrt{3}}{2}}+\frac{\sqrt{3}+1}{\frac{3-\sqrt{3}}{2}}=\frac{2\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{2\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}\)
\(=\frac{2}{\sqrt{3}}\left(\frac{4-2\sqrt{3}+4+2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right)=\frac{2}{\sqrt{3}}.\frac{8}{2}=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\)
\(\forall k\ge0\)ta có :
\(\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{k+1-k}=\sqrt{k+1}-\sqrt{k}\)
Bạn áp dụng công thức này vào dãy trên ta sẽ có các số hạng triệt tiêu đi nhau và ra kết quả
\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}=\sqrt{100}-1=10-1=9\)
A = \(\frac{1}{1+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\) + . . . . . . . . . + \(\frac{1}{\sqrt{99+\sqrt{100}}}\)
= \(\sqrt{2}\) - 1 + \(\sqrt{2}\) - \(\sqrt{3}\) + . . . . . . . + \(\sqrt{100}\) - \(\sqrt{99}\)
= - 1 + \(\sqrt{100}\) = \(\sqrt{100}\) - 1 = 10 - 1 = 9
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}\)
\(=\frac{\sqrt{3}.2}{\sqrt{3}}\)
\(=2\)
\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
\(=\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}+1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{1+\sqrt{3}}+1\right)}-\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}-1\right)}{\left(\sqrt{\sqrt{3}+1}+1\right)\left(\sqrt{1+\sqrt{3}}+1\right)}\)
\(=\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}+1\right)-\sqrt{3}.\left(\sqrt{1+\sqrt{3}}-1\right)}{\left(\sqrt{1+\sqrt{3}}-1\right)\left(\sqrt{1+\sqrt{3}}+1\right)}\)
\(=\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}+1\right)-\sqrt{3}.\left(\sqrt{1+\sqrt{3}}-1\right)}{\sqrt{3}}\)
\(=\frac{2\sqrt{3}}{\sqrt{3}}\)
= 2