Câu hỏi của Lê Thị Thanh Quỳnh

Question

Cho A=$\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$Chứng minh rằng:$\frac{1}{15}< A< \frac{1}{10}$

Phạm Tuấn Đạt · Accepted Answer

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$Câu trả lời: 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$

Fudo · Answer

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}$Câu trả lời: 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}$

Acsimet · Answer

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}$Câu trả lời: 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}$

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