Câu hỏi của nguyễn vũ kim anh

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1. Chứng minh :$\frac{5}{6}< \frac{1}{31}+\frac{1}{32}+...+\frac{1}{90}< \frac{3}{2}$ NHANH LÊN NHA MK CẦN GẤP GẤP LẮM LẮM LUÔN LUÔN ĐẤYAI NHANH MK CHO 1 TICK, LÚC 9h MK CÒN ĐI HOK THÊM NỮA

Lê Tài Bảo Châu · Accepted Answer

Đặt $A=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{90}$ $=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\right)+\left(\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}\right)$Đặt $B=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}$ Ta có: $\frac{1}{31}>\frac{1}{45}$ $\frac{1}{32}>\frac{1}{45}$ .................... $\frac{1}{45}=\frac{1}{45}$$\Rightarrow B>\frac{1}{45}.15$$\Rightarrow B>\frac{1}{3}$Đặt $C=\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}$Ta có: $\frac{1}{46}>\frac{1}{90}$ $\frac{1}{47}>\frac{1}{90}$ ..................... $\frac{1}{90}=\frac{1}{90}$$\Rightarrow C>\frac{1}{90}.45$$\Rightarrow C>\frac{1}{2}$$\Rightarrow B+C>\frac{1}{3}+\frac{1}{2}$Hay $A>\frac{5}{6}\left(1\right)$Lại có: $A=\left(\frac{1}{31}+...+\frac{1}{59}\right)+\left(\frac{1}{60}+...+\frac{1}{90}\right)$Đặt $D=\frac{1}{31}+...+\frac{1}{59}$Ta có: $\frac{1}{31}< \frac{1}{30}$ . ................... $\frac{1}{59}< \frac{1}{30}$$\Rightarrow D< \frac{1}{30}.60$$\Rightarrow D< \frac{1}{2}$Đăt $E=\frac{1}{60}+...+\frac{1}{90}$Ta có: $\frac{1}{60}=\frac{1}{60}$ ................. $\frac{1}{90}< \frac{1}{60}$$\Rightarrow E< \frac{1}{60}.31$$\Rightarrow E< \frac{31}{60}< 1$$\Rightarrow E< 1$$\Rightarrow E+D< 1+\frac{1}{2}$Hay $A< \frac{3}{2}\left(2\right)$Từ (1) và (2) $\Rightarrow\frac{5}{6}< A< \frac{3}{2}$Câu trả lời: Đặt $A=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{90}$ $=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\right)+\left(\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}\right)$Đặt $B=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}$ Ta có: $\frac{1}{31}>\frac{1}{45}$ $\frac{1}{32}>\frac{1}{45}$ .................... $\frac{1}{45}=\frac{1}{45}$$\Rightarrow B>\frac{1}{45}.15$$\Rightarrow B>\frac{1}{3}$Đặt $C=\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}$Ta có: $\frac{1}{46}>\frac{1}{90}$ $\frac{1}{47}>\frac{1}{90}$ ..................... $\frac{1}{90}=\frac{1}{90}$$\Rightarrow C>\frac{1}{90}.45$$\Rightarrow C>\frac{1}{2}$$\Rightarrow B+C>\frac{1}{3}+\frac{1}{2}$Hay $A>\frac{5}{6}\left(1\right)$Lại có: $A=\left(\frac{1}{31}+...+\frac{1}{59}\right)+\left(\frac{1}{60}+...+\frac{1}{90}\right)$Đặt $D=\frac{1}{31}+...+\frac{1}{59}$Ta có: $\frac{1}{31}< \frac{1}{30}$ . ................... $\frac{1}{59}< \frac{1}{30}$$\Rightarrow D< \frac{1}{30}.60$$\Rightarrow D< \frac{1}{2}$Đăt $E=\frac{1}{60}+...+\frac{1}{90}$Ta có: $\frac{1}{60}=\frac{1}{60}$ ................. $\frac{1}{90}< \frac{1}{60}$$\Rightarrow E< \frac{1}{60}.31$$\Rightarrow E< \frac{31}{60}< 1$$\Rightarrow E< 1$$\Rightarrow E+D< 1+\frac{1}{2}$Hay $A< \frac{3}{2}\left(2\right)$Từ (1) và (2) $\Rightarrow\frac{5}{6}< A< \frac{3}{2}$

Lê Tài Bảo Châu · Answer

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