Tinh:
\(1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}\)
K
Khách
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ML
25 tháng 6 2015
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 \(1-\frac{\sqrt{3}}{2}=\left(\frac{\sqrt{3}-1}{2}\right)^2\)
\(VT=\frac{\left(\frac{\sqrt{3}+1}{2}\right)^2}{1+\frac{\sqrt{3}+1}{2}}+\frac{\left(\frac{\sqrt{3}-1}{2}\right)^2}{1-\frac{\sqrt{3}-1}{2}}=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{3+\sqrt{3}}{2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{3-\sqrt{3}}{2}}\)\(=\frac{\left(\sqrt{3}+1\right)^2}{2.\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\left(\sqrt{3}-1\right)^2}{2.\sqrt{3}\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+1}{2\sqrt{3}}+\frac{\sqrt{3}-1}{2\sqrt{3}}=\frac{\sqrt{3}+1+\sqrt{3}-1}{2\sqrt{3}}=1=VP\)
1 tháng 8 2018
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)
\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
11 tháng 9 2017
\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
TQ
1
29 tháng 3 2018
= \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+ \(\frac{1}{15}\)
= \(\frac{10}{30}\)+ \(\frac{5}{30}\)+ \(\frac{3}{30}\)+ \(\frac{2}{30}\)= \(\frac{20}{30}\)= \(\frac{2}{3}\).(lấy mẫu chung là 30)
Học tốt !
ND
3
S
18 tháng 2 2019
\(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{99}}\)
\(2^2A=2+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}\)
\(3A=2-\frac{1}{2}=\frac{3}{2}\)
\(A=\frac{3}{2}\div3=\frac{1}{2}\)
MN
18 tháng 2 2019
A = 1/2 + 1/23 + 1/25 + .... + 1/299
1/22.A = 1/23 + 1/25 + 1/27 + .... + 1/2101
1/4.A - A = 1/23 + 1/25 + 1/27 + .... + 1/2101 - ( 1/2 + 1/23 + 1/25 + .... + 1/299 )
-3/4.A = 1/2101 - 1/2
3/4 .A = -(1/2101 - 1/2 )
A = (1/2101 + 1/2 )/3/4
Hok tốt !
6 tháng 9 2016
Xét mẫu số của F :
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2016\cdot2017}{2}}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{2017}\right)=2-\frac{2}{2017}=\frac{4032}{2017}\)
Suy ra : \(F=\frac{2.2016}{\frac{4032}{2017}}=\frac{2.2016.2017}{4032}=2017\)