cho M = 1/10 + 1/15 + 1/21 + 1/28 + ....+ 1/105 + 1 /120.Chung to 1/3 < M < 1/2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KK
19 tháng 4 2018
M=1/10 + 1/15 + 1/21 +....+ 1/120
M=2/20 +2/30+2/42+....+2/240
M=2/4.5 + 2/5.6 + 2/6.7 +.....+ 2/15.16
M=2.(1/4.5 +......+ 1/15.16)
M=2.(1/4 -1/5 +1/5 - 1/6 +.....+ 1/15 - 1/16)
M=2.(1/4 - 1/16)
M=2.(4/16 - 1/16)
M=2. 3/16
M=6/16=3/8
Có 1/3 = 8/24 < 9/24 = 3/8 =>1/3<M
Có 1/2 = 4/8>3/8 =>1/2 >M
=> 1/3 < M < 1/2
OY
27 tháng 7 2021
\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)
\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(M=2.\dfrac{3}{16}\)
\(M=\dfrac{3}{8}\)
Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)
NG
1
6 tháng 6 2019
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
Ta có :
\(\frac{1}{10}< 1\)
\(\frac{1}{15}< 1\)
\(\frac{1}{21}< 1\)
........................
\(\frac{1}{120}< 1\)
\(\Rightarrow\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}< 1\)
\(\Rightarrow A< 1\)( đpcm)
XO
6 tháng 6 2019
Ta có : A = \(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
= \(\frac{1}{20}\times2+\frac{1}{30}\times2+...+\frac{1}{240}\times2\)
= \(2\times\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
= \(2\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{16}\right)\)
= \(2\times\frac{3}{16}\)
= \(\frac{3}{8}\)< 1
=> A < 1
KT
9 tháng 8 2018
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{105}+\frac{1}{210}\)
=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{210}+\frac{1}{240}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{!}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
=> \(A=\frac{7}{8}\)
cho tiền rồi giải cho
\(M=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{105}+\frac{1}{120}\)
\(M=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{210}+\frac{2}{240}\)
\(M=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{14.15}+\frac{2}{15.16}\)
\(M=\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+...+\frac{2}{15}-\frac{2}{16}\)
\(M=\frac{2}{4}-\frac{2}{16}=\frac{3}{8}\)
Vì \(\frac{3}{9}< \frac{3}{8}< \frac{4}{8}\)nên \(\frac{1}{3}< M< \frac{1}{2}\)
Vậy \(\frac{1}{3}< M< \frac{1}{2}\)
P/S : Đừng nói như lần trước nhé!