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hãy nhìn kĩ hihi
vì mỗi p/số của M đều bé hơn 1,áp dụng quy tắc thứ 7 để so sánh có
1/2<1+1/2+1=2/3(xảy ra khi p/số<1 như trên)
3/4<3+1/4+1=4/5
.......
99/100<99+1/100+1=100/101
tích chúng sẽ bé hơn
2/3.4/5.6/7......100/101=N
Vậy M<N
M.N=1/2.2/3.3/4.......99/100.100/101
tử và mẫu xuất hiện số đối nhau,khử đi còn
M.N=1/101
Dựa vào câu a,b có
M.M<M.N(vì N>M)
M.M<1/101
dpcm là M<1/10
M.M<1/10.1/10=1/100
mà M^2<1/101<1/100=1/10^2
=>M<1/10
hơi vắt óc nên xin olm tích cho nha
chúc học tốt
c) \(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{1}{2}.\frac{4}{4}.\frac{6}{6}...\frac{100}{100}=\frac{1}{2}\)
giải tương tự như câu hôm qua mình giải
để chứng minh A < \(\frac{1}{10}\). Ta thấy \(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow A^2< \left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(=\frac{1.\left(3.5...99\right)}{2.4.6...100}.\frac{2.4.6...100}{\left(3.5.7...99\right).101}\)
\(=\frac{1}{101}< \frac{1}{10}\)
\(\Rightarrow A^2< \frac{1}{101}< \frac{1}{100}=\frac{1}{10^2}\Rightarrow A< \frac{1}{10}\)
để chứng minh A > \(\frac{1}{15}\). Ta thấy \(A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\right)\)
\(=\frac{1.\left(3.5...99\right)}{\left(2.4.6...98\right).100}.\frac{1.\left(2.4...98\right)}{2.\left(3.5...99\right)}\)
\(=\frac{1}{100}.\frac{1}{2}=\frac{1}{200}\)
\(\Rightarrow A^2>\frac{1}{200}>\frac{1}{225}=\frac{1}{15^2}\Rightarrow A>\frac{1}{15}\)
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 = \(\frac{1}{101}\)\(< \frac{1}{100}\)
\(\Rightarrow M< \frac{1}{10}\)
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
\(\frac{1}{2}< \frac{2}{3}\text{ ; }\frac{3}{4}< \frac{4}{5}\text{ ; }\frac{5}{6}< \frac{6}{7}\text{ ; }...\text{ ; }\frac{99}{100}< \frac{100}{101}\)
\(\text{Đặt }B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A\cdot A< A\cdot B=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right)\)
\(A\cdot A< A\cdot B=\frac{1}{101}< \frac{1}{10}\)
\(A^2< \frac{1}{10}\text{ }\Rightarrow\text{ }A< \frac{1}{10}^{^{\left(1\right)}}\)
\(\frac{1}{2}=\frac{1}{2}\text{ ; }\frac{3}{4}>\frac{2}{3}\text{ ; }\frac{5}{6}>\frac{4}{5}\text{ ; }...\text{ ; }\frac{99}{100}>\frac{98}{99}\)
\(\text{Đặt }C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)
\(A\cdot C=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\right)\)
\(A\cdot C=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)
\(A\cdot C=\frac{1}{200}\)
\(\text{Vì }A>C\text{ }\Rightarrow\text{ }A^2>A\cdot C=\frac{1}{200}\)
\(\text{Mà }A^2>\frac{1}{200}>\frac{1}{15^2}\)
\(\Rightarrow\text{ }A>\frac{1}{15}^{^{\left(2\right)}}\)
\(\text{Từ }^{\left(1\right)}\text{ và }^{\left(2\right)}\)
\(\Rightarrow\text{ }\frac{1}{15}< A< \frac{1}{10}\)
\(\Rightarrow\text{ }\text{ĐPCM}\)
