Cho B=1/2-1/3+1/4-1/5+...-1/99. C/m B<2/5
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PT
3 tháng 3 2017
Bài 1:
\(\dfrac{5}{x} - \dfrac{y}{3} =\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{y}{3}=\dfrac{5}{x}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{2y}{6}=\dfrac{5}{x}\)
\(\Rightarrow1+\dfrac{2y}{6}=\dfrac{5}{x}\)
\(\Rightarrow x.\left(1+2y\right)=30\)
Vì \(2y\) chẵn nên \(1+2y\) lẻ
\(\Rightarrow1+2y\in\left\{\pm1;\pm3;\pm5;\pm30\right\}\)
\(\Rightarrow x\in\left\{\pm10;\pm30;\pm6;\pm2\right\}\)
PT
3 tháng 3 2017
Bài 2:
\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{\left(2n-2\right).2n}\)
\(=\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{\left(2n-2\right).2n}\right).\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)
\(=\dfrac{1}{4}-\dfrac{1}{2n.2}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(đpcm\right)\)
HH
23 tháng 6 2018
Bài 2:
\(E=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(\Leftrightarrow E=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)
\(\Leftrightarrow E=\dfrac{3.4.5...100}{2.3.4...99}\)
\(\Leftrightarrow E=\dfrac{\left(3.4.5...99\right).100}{2.\left(3.4...99\right)}\)
\(\Leftrightarrow E=\dfrac{100}{2}=50\)
Vậy ...
PH
cho:
m = 1/2*3/4*5/6*....*99/100
n = 2/3*4/5*6/7*...*100/101
a, Chứng tỏ m<n
b,Tìm m*n
c, chứng tỏ m<1/10
0
30 tháng 8 2020
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)