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Question

CM rằng $\frac{1}{6}< \frac{1}{5x5}+\frac{1}{6x6}+\frac{1}{7x7}+............+\frac{1}{100x100}< \frac{1}{4}$

๛Ňɠũ Vị Čáէツ · Accepted Answer

Ta có:1/5×5 < 1/4×51/6×6 < 1/5×61/7×7 < 1/6×7.........1/100×100 < 1/99×100=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 < 1/4×5 + 1/5×6 + 1/6×7 +.....+ 1/99×100 = 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +......+ 1/99-1/100 = 1/4-1/100 < 1/4 => 1/5×5 + 1/6×6+1/7×7 +...+1/100×100<1/4 (1)Lại có:1/5×5 > 1/6×71/6×6 > 1/7×81/7×7 > 1/8×9........1/100×100 > 1/101×102=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 > 1/5×6 + 1/6×7 + 1/7×8 +.....+1/100×101 = 1/5-1/6 + 1/6-1/7 + 1/7-1/8 +.....+ 1/100 - 1/101 = 1/5 - 1/101 > 1/5 - 1/30 = 1/6=> 1/5×5 + 1/6×6 +1/7×7 +.....+ 1/100×100>1/6 (2)Từ (1) và (2)=> 1/6 < 1/5×5 +1/6×6+ 1/7×7 +...+1/100×100<1/4Câu trả lời: Ta có:1/5×5 < 1/4×51/6×6 < 1/5×61/7×7 < 1/6×7.........1/100×100 < 1/99×100=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 < 1/4×5 + 1/5×6 + 1/6×7 +.....+ 1/99×100 = 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +......+ 1/99-1/100 = 1/4-1/100 < 1/4 => 1/5×5 + 1/6×6+1/7×7 +...+1/100×100<1/4 (1)Lại có:1/5×5 > 1/6×71/6×6 > 1/7×81/7×7 > 1/8×9........1/100×100 > 1/101×102=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 > 1/5×6 + 1/6×7 + 1/7×8 +.....+1/100×101 = 1/5-1/6 + 1/6-1/7 + 1/7-1/8 +.....+ 1/100 - 1/101 = 1/5 - 1/101 > 1/5 - 1/30 = 1/6=> 1/5×5 + 1/6×6 +1/7×7 +.....+ 1/100×100>1/6 (2)Từ (1) và (2)=> 1/6 < 1/5×5 +1/6×6+ 1/7×7 +...+1/100×100<1/4

Phạm Tuấn Đạt · Answer

Đặt $A=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}$Có $\frac{1}{5.5}< \frac{1}{4.5};\frac{1}{6.6}< \frac{1}{5.6};...;\frac{1}{100.100}< \frac{1}{99.100}$$\Rightarrow A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}$(1)Lại có :$\frac{1}{5.5}>\frac{1}{5.6};\frac{1}{6.6}>\frac{1}{6.7};...;\frac{1}{100.100}>\frac{1}{100.101}$$\Rightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\left(2\right)$Từ (1) và (2) $\RightarrowĐCCM$Câu trả lời: Đặt $A=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}$Có $\frac{1}{5.5}< \frac{1}{4.5};\frac{1}{6.6}< \frac{1}{5.6};...;\frac{1}{100.100}< \frac{1}{99.100}$$\Rightarrow A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}$(1)Lại có :$\frac{1}{5.5}>\frac{1}{5.6};\frac{1}{6.6}>\frac{1}{6.7};...;\frac{1}{100.100}>\frac{1}{100.101}$$\Rightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\left(2\right)$Từ (1) và (2) $\RightarrowĐCCM$

Kiên-Messi-8A-Boy2k6 · Answer

Đặt $X=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}$$\Rightarrow X< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}$$\Rightarrow X=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$$\Rightarrow X< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}$$\Rightarrow X< \frac{1}{4}\left(1\right)$$\Rightarrow X>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}$$\Rightarrow X>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}$$\Rightarrow X>\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}$$\Rightarrow X>\frac{1}{6}\left(2\right)$Từ $\left(1\right);\left(2\right)\Rightarrow\frac{1}{6}< X< \frac{1}{4}$Câu trả lời: Đặt $X=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}$$\Rightarrow X< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}$$\Rightarrow X=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$$\Rightarrow X< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}$$\Rightarrow X< \frac{1}{4}\left(1\right)$$\Rightarrow X>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}$$\Rightarrow X>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}$$\Rightarrow X>\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}$$\Rightarrow X>\frac{1}{6}\left(2\right)$Từ $\left(1\right);\left(2\right)\Rightarrow\frac{1}{6}< X< \frac{1}{4}$

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