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2 tháng 9 2017
Ta có : \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};....;\frac{99}{100}< \frac{100}{101}\)
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)\(\Rightarrow B>A\)
\(\Rightarrow A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(\Rightarrow A.B=\frac{1}{101}\)
Vì \(B>A\)\(\Rightarrow A.B>A.A=A^2\)
\(\Rightarrow\frac{1}{101}>A^2\)
Mà \(\frac{1}{10^2}>\frac{1}{101}>A^2\Rightarrow\frac{1}{10^2}>A^2\)
\(\Rightarrow\frac{1}{10}< A\left(1\right)\)\(\)
Ta lai có :
\(\frac{1}{2}=\frac{1}{2};\frac{3}{4}>\frac{2}{3};\frac{5}{6}>\frac{4}{5};...;\frac{99}{100}>\frac{98}{99}\)
Đặt \(C=\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A.C=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\right)\)
\(\Rightarrow A.C=\frac{1}{2}.\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A.C=\frac{1}{200}\)
Vì \(A>C\)
\(\Rightarrow A^2>A.C=\frac{1}{200}\)
Mà \(A^2>\frac{1}{200}>\frac{1}{15^2}\)
\(\Rightarrow A^2>\frac{1}{15^2}\)
\(\Rightarrow A>\frac{1}{15}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)
\(\Rightarrow\frac{1}{15}< A< \frac{1}{10}\)
\(\RightarrowĐPCM\)
F
26 tháng 5 2019
Bài giải
\(\frac{1}{2}< \frac{2}{3}\text{ ; }\frac{3}{4}< \frac{4}{5}\text{ ; }\frac{5}{6}< \frac{6}{7}\text{ ; }...\text{ ; }\frac{99}{100}< \frac{100}{101}\)
\(\text{Đặt }B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A\cdot A< A\cdot B=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right)\)
\(A\cdot A< A\cdot B=\frac{1}{101}< \frac{1}{10}\)
\(A^2< \frac{1}{10}\text{ }\Rightarrow\text{ }A< \frac{1}{10}^{^{\left(1\right)}}\)
\(\frac{1}{2}=\frac{1}{2}\text{ ; }\frac{3}{4}>\frac{2}{3}\text{ ; }\frac{5}{6}>\frac{4}{5}\text{ ; }...\text{ ; }\frac{99}{100}>\frac{98}{99}\)
\(\text{Đặt }C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)
\(A\cdot C=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\right)\)
\(A\cdot C=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)
\(A\cdot C=\frac{1}{200}\)
\(\text{Vì }A>C\text{ }\Rightarrow\text{ }A^2>A\cdot C=\frac{1}{200}\)
\(\text{Mà }A^2>\frac{1}{200}>\frac{1}{15^2}\)
\(\Rightarrow\text{ }A>\frac{1}{15}^{^{\left(2\right)}}\)
\(\text{Từ }^{\left(1\right)}\text{ và }^{\left(2\right)}\)
\(\Rightarrow\text{ }\frac{1}{15}< A< \frac{1}{10}\)
\(\Rightarrow\text{ }\text{ĐPCM}\)
S
18 tháng 5 2017
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
26 tháng 3 2016
mk cm vế sau, vế đầu tương tự nha:
ta có:
1/2<2/3
3/4<4/5
...
99/100<100/101
=>A^2<1/2×2/3×3/4×...×99/100×100/101
=>A^2<1/101<1/100
=>A<1/10(dpcm)
PX
6 tháng 4 2017
a, ta xét:
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.....
\(\frac{99}{100}< \frac{100}{101}\)
=>\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)
hay:A<B(đpcm)
b,\(A.B=\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.....\frac{100}{101}\)
\(=\frac{1.2.3....100}{2.3.4....101}=\frac{1}{101}\)
c,vì A<B (theo phần a)
=>A.A<B.A
Mà B.A=\(\frac{1}{101}\)
=>A2<101
Mà A2=\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)
=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{101}\)<\(\frac{1}{100}=\frac{1}{10^2}\)
=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{10^2}\)
=>\(\frac{1}{2}.\frac{3}{4}....\frac{99}{100}< \frac{1}{10}\)
Hay A<\(\frac{1}{10}\)
Cái này mình tìm ở trên Google (trang web: Lazi. URL: https://lazi.vn/edu/exercise/2456647/cho-a-1-32-3-34-5-36-99-3100-chung-minh-a-be-hon-5-32)😊(●'◡'●)
Bạn vào coi nha.