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\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}.\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}\)\(+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}\)
\(=\frac{\frac{4+2\sqrt{3}}{4}}{1+\sqrt{\frac{4+\sqrt{3}}{4}}}\)\(+\frac{\frac{4-2\sqrt{3}}{4}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}}\)
\(=\frac{\frac{3+2\sqrt{3}+1}{4}}{1+\sqrt{\frac{3+2\sqrt{3}+1}{4}}}\)\(+\frac{\frac{3-2\sqrt{3}+1}{4}}{1-\sqrt{\frac{3-2\sqrt{3}+1}{4}}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{2+\sqrt{3}}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{2-\sqrt{3}}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{\left(\sqrt{3}+1\right)^2}{4}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{\left(\sqrt{3}-1\right)^2}{4}}\)
\(=1+1=2\)
\(A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(A=\frac{2\left(1+\frac{\sqrt{3}}{2}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\left(1-\frac{\sqrt{3}}{2}\right)}{2-\sqrt{4-2\sqrt{3}}}\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(A=\frac{\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}\)
\(A=\frac{3+\sqrt{3}+3-\sqrt{3}}{6}\)
\(A=\frac{6}{6}=1\)
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
a, bạn chỉ cần lập công thức tông quát :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cái này bạn chỉ cần trục căn thức ở mẫu chưng minh xong áp dụng vào luôn là ra
a, kq : 4/5
b,\(1-\frac{1}{\sqrt{n+1}}\)
c,d chưa nghĩ ra
ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{\left(n+1\right)n}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{\left(n+1\right)n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
nên: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}=\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)\(=1-\frac{1}{5}=\frac{4}{5}\)
Ta có: \(4\left(1+\frac{\sqrt{3}}{2}\right)=3+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\Rightarrow1+\frac{\sqrt{3}}{2}=\left(\frac{\sqrt{3}+1}{2}\right)^2\)
Tương tự \(1-\frac{\sqrt{3}}{2}=\left(\frac{\sqrt{3}-1}{2}\right)^2\)
\(VT=\frac{\left(\frac{\sqrt{3}+1}{2}\right)^2}{1+\frac{\sqrt{3}+1}{2}}+\frac{\left(\frac{\sqrt{3}-1}{2}\right)^2}{1-\frac{\sqrt{3}-1}{2}}=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{3+\sqrt{3}}{2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{3-\sqrt{3}}{2}}\)\(=\frac{\left(\sqrt{3}+1\right)^2}{2.\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\left(\sqrt{3}-1\right)^2}{2.\sqrt{3}\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+1}{2\sqrt{3}}+\frac{\sqrt{3}-1}{2\sqrt{3}}=\frac{\sqrt{3}+1+\sqrt{3}-1}{2\sqrt{3}}=1=VP\)