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\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
Chúc bạn học tốt ~
Số lượng số của dãy số trên là :
( 80 - 41 ) : 1 + 1 = 40 ( số )
Ta có :
\(\frac{1}{41}>\frac{1}{80};\frac{1}{42}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\)( 40 số )
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{80}.40=\frac{1}{2}\left(1\right)\)
Ta có :
\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{80}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( 40 số )
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{40}.40=1\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{2}< \frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}< 1\left(Đpcm\right)\)
Chúc bạn học tốt !!!
Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12
Bài 1:
\(\dfrac{5}{x} - \dfrac{y}{3} =\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{y}{3}=\dfrac{5}{x}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{2y}{6}=\dfrac{5}{x}\)
\(\Rightarrow1+\dfrac{2y}{6}=\dfrac{5}{x}\)
\(\Rightarrow x.\left(1+2y\right)=30\)
Vì \(2y\) chẵn nên \(1+2y\) lẻ
\(\Rightarrow1+2y\in\left\{\pm1;\pm3;\pm5;\pm30\right\}\)
\(\Rightarrow x\in\left\{\pm10;\pm30;\pm6;\pm2\right\}\)
Bài 2:
\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{\left(2n-2\right).2n}\)
\(=\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{\left(2n-2\right).2n}\right).\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)
\(=\dfrac{1}{4}-\dfrac{1}{2n.2}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(đpcm\right)\)
Ta có: \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{40}.20=\frac{1}{2}\)
\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{1}{60}.20=\frac{1}{3}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}< \frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{3+2}{6}=\frac{5}{6}\) (đpcm)
Bài 1:
Ta có: \(\frac{1}{51}>\frac{1}{100}\)
\(\frac{1}{52}>\frac{1}{100}\)
......
\(\frac{1}{99}>\frac{1}{100}\)
Công vế với vế lại ta được:
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\) (1)
Lại có: \(\frac{1}{51}< \frac{1}{50}\)
\(\frac{1}{52}< \frac{1}{50}\)
.....
\(\frac{1}{100}< \frac{1}{50}\)
Cộng vế với vế lại ta được:
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{50}{50}=1\) (2)
Từ (1)(2) => \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< 1\) (đpcm)
Bài 2:
Đặt S = 1/41 + 1/42 +...+ 1/80
S có 40 số hạng,chia thành 4 nhóm,mỗi nhóm có 10 số hạng
Ta có:S = \(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\) + \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)+ \(\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)\)+ \(\left(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}\right)\)
=> S > \(\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\right)+\left(\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\right)\)
=> S > \(\frac{10}{50}+\frac{10}{60}+\frac{10}{70}+\frac{10}{80}\)
=> S > \(\frac{533}{840}>\frac{490}{840}=\frac{7}{12}\)
Vậy \(S=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{7}{12}\left(đpcm\right)\)