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\(=\frac{\sqrt{\frac{2+2\sqrt{2}+1}{3}}+\sqrt{\frac{2-2\sqrt{2}+1}{3}}}{\sqrt{\frac{2+2\sqrt{2}+1}{3}}-\sqrt{\frac{2-2\sqrt{2}+1}{3}}}\)
\(=\frac{\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{3}}+\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{3}}}{\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{3}}-\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{3}}}\)
\(=\frac{\frac{\sqrt{2}+1+\sqrt{2}-1}{\sqrt{3}}}{\frac{\sqrt{2}+1-\sqrt{2}+1}{\sqrt{3}}}=\frac{\frac{2\sqrt{2}}{\sqrt{3}}}{\frac{2}{\sqrt{3}}}=\sqrt{2}\)
\(=\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}=\frac{\sqrt{2}+1+\sqrt{2}-1}{\sqrt{2}+1-\sqrt{2}+1}=\frac{2\sqrt{2}}{2}=\sqrt{2}\)
Nhan \(\sqrt{\frac{2}{2}}\) vao hai ve cua bieu thuc ta duoc
\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3}+1}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}+1}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\)
Toi day quy dong thoi minh lam nhanh nha
\(=\frac{\sqrt{6}+3\sqrt{2}-\sqrt{6}+3\sqrt{2}}{6}\)
\(=\frac{6\sqrt{2}}{6}=\sqrt{2}\)
TIck cho mifnh nha
Đặt \(A=\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}{\sqrt{2}\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}+1-\sqrt{3}+1}=\frac{2\sqrt{3}}{2}=\sqrt{3}\)
Vậy \(A=\sqrt{3}\)
Với \(k\in N;k\ne0\) ta có :
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{\left(k+1\right)}}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}\)
\(=\frac{\sqrt{k+1}+\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\)
\(=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng ta có :
\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
0.6636458856^v^