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Ta có: \(S=\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}\). Theo như quy tắc đã học ở lớp 5. Ta có:
Các phân số có tử bé hơn mẫu thì phân số đó bé hơn 1
Mà \(\frac{6}{15};\frac{6}{16};\frac{6}{17};\frac{6}{18};\frac{6}{19}\) đều bé hơn 1.
\(\Rightarrow\frac{6}{15}+\frac{6}{16}+\frac{6}{17}+\frac{6}{18}+\frac{6}{19}< 0\RightarrowĐPCM\) (Vì: \(1>\frac{6}{15}>\frac{6}{16}>\frac{6}{17}>\frac{6}{18}>\frac{6}{19}\))
ta có:
6/15+6/16+6/17+6/18+6/19
=31/40+6/17+6/18+6/19
=767/680+6/18+6/19
=1.7777
vậy s không thuộc n
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
\(\frac{1}{2}< \frac{2}{3}\text{ ; }\frac{3}{4}< \frac{4}{5}\text{ ; }\frac{5}{6}< \frac{6}{7}\text{ ; }...\text{ ; }\frac{99}{100}< \frac{100}{101}\)
\(\text{Đặt }B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A\cdot A< A\cdot B=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right)\)
\(A\cdot A< A\cdot B=\frac{1}{101}< \frac{1}{10}\)
\(A^2< \frac{1}{10}\text{ }\Rightarrow\text{ }A< \frac{1}{10}^{^{\left(1\right)}}\)
\(\frac{1}{2}=\frac{1}{2}\text{ ; }\frac{3}{4}>\frac{2}{3}\text{ ; }\frac{5}{6}>\frac{4}{5}\text{ ; }...\text{ ; }\frac{99}{100}>\frac{98}{99}\)
\(\text{Đặt }C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)
\(A\cdot C=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\right)\)
\(A\cdot C=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)
\(A\cdot C=\frac{1}{200}\)
\(\text{Vì }A>C\text{ }\Rightarrow\text{ }A^2>A\cdot C=\frac{1}{200}\)
\(\text{Mà }A^2>\frac{1}{200}>\frac{1}{15^2}\)
\(\Rightarrow\text{ }A>\frac{1}{15}^{^{\left(2\right)}}\)
\(\text{Từ }^{\left(1\right)}\text{ và }^{\left(2\right)}\)
\(\Rightarrow\text{ }\frac{1}{15}< A< \frac{1}{10}\)
\(\Rightarrow\text{ }\text{ĐPCM}\)
a) Ta có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}>\frac{6}{19}.5=\frac{30}{19}>1\)
\(\Rightarrow S>1\)
Ta lại có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}< \frac{6}{15}.5=\frac{30}{15}=2\)
\(\Rightarrow S< 2\)
Vậy, 1 < S < 2
b) \(1< S< 2\Rightarrow S\notin Z\)