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Đặt \(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{60}\right)+...+\frac{1}{70}\)
Nhận xét:
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+...+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}\)
\(\Rightarrow A>\frac{4}{3}\)
Vậy \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}>\frac{4}{3}\) (Đpcm)
\(A=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+....+\dfrac{1}{70}\\ =\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\right)+\left(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}\right)+\left(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}\right)+....+\dfrac{1}{70}\\ \)
\(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}=\dfrac{1}{2}\)
\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+....+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)
\(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\)
\(\Rightarrow A>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{61}+...+\dfrac{1}{70}>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{4}{3}\)
Chúc bạn học tốt !!!!!!
Ta có: \(\dfrac{3}{10}>\dfrac{3}{15}\)
\(\dfrac{3}{11}>\dfrac{3}{15}\)
\(\dfrac{3}{12}>\dfrac{3}{15}\)
\(\dfrac{3}{13}>\dfrac{3}{15}\)
\(\dfrac{3}{14}>\dfrac{3}{15}\)
Do đó: \(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}=1\)
hay 1<S(1)
Ta có: \(\dfrac{3}{11}< \dfrac{3}{10}\)
\(\dfrac{3}{12}< \dfrac{3}{10}\)
\(\dfrac{3}{13}< \dfrac{3}{10}\)
\(\dfrac{3}{14}< \dfrac{3}{10}\)
Do đó: \(\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}=\dfrac{12}{10}\)
\(\Leftrightarrow S< \dfrac{15}{10}=\dfrac{3}{2}< 2\)(2)
Từ (1) và (2) suy ra 1<S<2(đpcm)
Lời giải:
Ta có:
$\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}=\frac{10}{11}<1$
Ta có điều phải chứng minh
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
\(\dfrac{5}{2x1}+\dfrac{4}{1x11}+\dfrac{3}{11x2}+\dfrac{1}{2x15}+\dfrac{13}{15x4}+\dfrac{15}{4x13}\)
=7x(\(\dfrac{5}{2x7}+\dfrac{4}{7x11}+\dfrac{3}{11x14}+\dfrac{1}{14x15}+\dfrac{13}{15x28}+\dfrac{15}{28x43}\))
=7x\(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}+\dfrac{1}{28}-\dfrac{1}{43}\)=7x(\(\dfrac{1}{2}-\dfrac{1}{43}\))
=7x\(\dfrac{41}{86}\)
=\(\dfrac{287}{86}\)
5/2x1+4/1x11+3/11x2+1/2x15+13/15x4+15/4x43=7x(5/2x7+4/7x11+3/11x14+1/14x15+13/15x28+15/28x43)=7x(1/2-1/7+1/7-1/11+1/11-1/14+1/14+1/15+1/15-1/28+1/28-1/43)=7x(1/2-1/43)=7x41/86=287/86
bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
