Cho A=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}...+\frac{1}{5^{500}}chứngminhA< \frac{1}{4}\)
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NJ
27 tháng 3 2017
\(A=1+\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+..........+\frac{1}{50.50}\)
\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{49}{50}=1\frac{49}{50}\)
vì\(1\frac{49}{50}< 2\Rightarrow A< 2\)
9 tháng 2 2017
\(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{\frac{21}{44}+\frac{3}{13}}{\frac{20}{77}+\frac{5}{13}}+\frac{\frac{1}{6}+\frac{1}{4}}{\frac{5}{12}+\frac{5}{8}}\)
\(=\frac{\frac{405}{572}}{\frac{645}{1001}}+\frac{\frac{5}{12}}{\frac{25}{24}}\)
\(=\frac{1289}{860}\)
24 tháng 4 2018
A =\(\frac{\left(\frac{17}{5}+\frac{1}{5}\right).\frac{2}{5}}{\left(\frac{38}{7}-\frac{9}{4}\right).\frac{56}{267}}\)
A=\(\frac{36}{25}\).\(\frac{3}{2}\)=\(\frac{54}{25}\)=2,16
B=\(\frac{1,2:\left(\frac{6}{5}-\frac{5}{4}\right)}{0,32+\frac{2}{25}}\)=-24.\(\frac{5}{2}\)=-60
vì 2,16 > -60 Vậy A>B
Ta có \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{500}}\)
=> 5A = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{499}}\)
Khi đó 5A - A = \(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{499}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{500}}\right)\)
=> 4A = \(1-\frac{1}{5^{500}}\)
=> \(A=\frac{1}{4}-\frac{1}{4.5^{500}}< \frac{1}{4}\)
=> A < 1/4 (đpcm)
A=1/5+1/5^2+1/5^3+1/5^4+...+1/5^500
5A=1+1/5+1/5^2+1/5^3+...+1/5^499
5A-A=(1+1/5+1/5^2+1/5^3+...+1/5^499)-(1/5+1/5^2+1/5^3+1/5^4+...+1/5^500)
4A=1-1/5^500
A=(1-1/5^500)/4<1/4
A<1/4