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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 \(\left(1\right);\left(2\right)\)
\(\Rightarrow\frac{1}{15}< A< \frac{1}{10}\)
\(\RightarrowĐPCM\)
Bài giải
\(\frac{1}{2}< \frac{2}{3}\text{ ; }\frac{3}{4}< \frac{4}{5}\text{ ; }\frac{5}{6}< \frac{6}{7}\text{ ; }...\text{ ; }\frac{99}{100}< \frac{100}{101}\)
\(\text{Đặt }B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A\cdot A< A\cdot B=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right)\)
\(A\cdot A< A\cdot B=\frac{1}{101}< \frac{1}{10}\)
\(A^2< \frac{1}{10}\text{ }\Rightarrow\text{ }A< \frac{1}{10}^{^{\left(1\right)}}\)
\(\frac{1}{2}=\frac{1}{2}\text{ ; }\frac{3}{4}>\frac{2}{3}\text{ ; }\frac{5}{6}>\frac{4}{5}\text{ ; }...\text{ ; }\frac{99}{100}>\frac{98}{99}\)
\(\text{Đặt }C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)
\(A\cdot C=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\right)\)
\(A\cdot C=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)
\(A\cdot C=\frac{1}{200}\)
\(\text{Vì }A>C\text{ }\Rightarrow\text{ }A^2>A\cdot C=\frac{1}{200}\)
\(\text{Mà }A^2>\frac{1}{200}>\frac{1}{15^2}\)
\(\Rightarrow\text{ }A>\frac{1}{15}^{^{\left(2\right)}}\)
\(\text{Từ }^{\left(1\right)}\text{ và }^{\left(2\right)}\)
\(\Rightarrow\text{ }\frac{1}{15}< A< \frac{1}{10}\)
\(\Rightarrow\text{ }\text{ĐPCM}\)
gọi biểu thức cần CM là B
\(3B=1+\frac23+\frac{3}{2^2}+\frac{4}{3^3}+\cdots+\frac{100}{3^{99}}\)
=> \(3B-B=1+\left(\frac23-\frac13\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+\cdots+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)-\frac{100}{3^{100}}\)
\(2B=1+\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
đặt C= \(\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}\)
=> \(3C=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\)
=> \(3C-C=\left(1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\right)-\left(\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}\)
=> \(C=\frac12-\frac{1}{2\cdot3^{99}}\)
\(2B=1+\frac12-\left(\frac{1}{2\cdot3^{99}}+\frac{100}{3^{100}}\right)\)
vì trong ngoặc lớn hơn 0
=> \(2B<\frac32\)
\(B<\frac34\left(đpcm\right)\)
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\left(\frac{1}{4\cdot5}+.\ldots+\frac{1}{99\cdot100}\right)\)
\(A=\frac{7}{12}+\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)\)
Mà \(\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)>0\)
=> A>\(\frac{7}{12}\)
mặt khác ta có: \(A=1-\frac12+\frac13-\frac14+\frac14-\frac15+\cdots+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\left(\frac12-\frac13\right)-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)
\(A=\frac56-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)
=> \(A<\frac56\)
Vậy \(\frac{7}{12}<A<\frac56\)
Sửa N=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)
Ta có : \(\frac{1}{2}< \frac{2}{3}\); \(\frac{3}{4}< \frac{4}{5}\); \(\frac{5}{6}< \frac{6}{7}\); ... ; \(\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)hay M < N
b) M .N = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}=\frac{1.2.3.4.5.6...99.100}{2.3.4.5.6.7...100.101}=\frac{1}{101}\)
c) vì M < N nên M. M < M . N = \(\frac{1}{101}\)\(< \frac{1}{100}\)
\(\Rightarrow M< \frac{1}{10}\)
vì 1/5 > 1/10 nên biểu thức trên sai
Vì \(\frac{1}{5}>\frac{1}{10}\)
=>vô lý
Vậy bất đảng thức \(\frac{1}{5}<\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}<\frac{1}{10}\) là sai
k nha