Chứng minh
\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{1}{10}\)
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gọi 2/3.4/5.6/7.....100/101=B
ta thấy:\(\frac{2}{3}>\frac{1}{2}\)
\(\frac{4}{5}>\frac{3}{4}\)
\(\frac{6}{7}>\frac{5}{6}\)
.....
\(\frac{100}{101}>\frac{99}{100}\)
\(\Rightarrow\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}>\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)mà:
Y.B=1/2.3/4.5/6.....99/100.2/3.4/5.6/7.....100/101
Y.B=\(\frac{1.3.5.....99.2.4.6....100}{2.4.6....100.3.5.7....101}\)
Y.B=\(\frac{1}{101}\)
vì Y<B \(\Rightarrow\)Y.B>Y.Y
=>Y.Y<\(\frac{1}{101}<\frac{1}{100}\)
=>Y.Y<\(\frac{1}{10}.\frac{1}{10}\)
=>y<1/10(đpcm)
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
\(\frac{1}{2}< \frac{2}{3}\text{ ; }\frac{3}{4}< \frac{4}{5}\text{ ; }\frac{5}{6}< \frac{6}{7}\text{ ; }...\text{ ; }\frac{99}{100}< \frac{100}{101}\)
\(\text{Đặt }B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(\Rightarrow\text{ }A\cdot A< A\cdot B=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right)\)
\(A\cdot A< A\cdot B=\frac{1}{101}< \frac{1}{10}\)
\(A^2< \frac{1}{10}\text{ }\Rightarrow\text{ }A< \frac{1}{10}^{^{\left(1\right)}}\)
\(\frac{1}{2}=\frac{1}{2}\text{ ; }\frac{3}{4}>\frac{2}{3}\text{ ; }\frac{5}{6}>\frac{4}{5}\text{ ; }...\text{ ; }\frac{99}{100}>\frac{98}{99}\)
\(\text{Đặt }C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)
\(A\cdot C=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\right)\)
\(A\cdot C=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)
\(A\cdot C=\frac{1}{200}\)
\(\text{Vì }A>C\text{ }\Rightarrow\text{ }A^2>A\cdot C=\frac{1}{200}\)
\(\text{Mà }A^2>\frac{1}{200}>\frac{1}{15^2}\)
\(\Rightarrow\text{ }A>\frac{1}{15}^{^{\left(2\right)}}\)
\(\text{Từ }^{\left(1\right)}\text{ và }^{\left(2\right)}\)
\(\Rightarrow\text{ }\frac{1}{15}< A< \frac{1}{10}\)
\(\Rightarrow\text{ }\text{ĐPCM}\)
Bạn tham khảo ở link này nhé :
Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến
ko can phai tinh ma la phan tich a vua co 1/15 vua co 1/10 thi co the so sanh ok?
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)
đặt \(A=\frac{1}{2}.\frac{3}{4}..\frac{99}{100}\)
ta có: \(A^2=\frac{1}{2}.\frac{1}{2}.\frac{3}{4}.\frac{3}{4}...\frac{99}{100}.\frac{99}{100}\)
\(\le\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}.\frac{100}{101}=\frac{1}{101}< \frac{1}{100}\)
\(A^2< \frac{1}{100}\Rightarrow A< \frac{1}{10}\)