Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)
\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(M=2.\dfrac{3}{16}\)
\(M=\dfrac{3}{8}\)
Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)
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
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 +...+ 1/14 - 1/15 + 1/15 - 1/16
= 1/4 - 1/16
= 3/16
Công thức : 1/ a x (a+1) = 1/a - 1/ a+1
\(M=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{105}+\frac{1}{120}\)
\(M=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{210}+\frac{2}{240}\)
\(M=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{14.15}+\frac{2}{15.16}\)
\(M=\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+...+\frac{2}{15}-\frac{2}{16}\)
\(M=\frac{2}{4}-\frac{2}{16}=\frac{3}{8}\)
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!
1/3.4 + 1/4.5 + 1/5.6+ 1/6.7+....+1/x(x+1) =3/10
1/3 -1/4 + 1/4-1/5+ 1/5 -1/6+......+1/x -1/x+1 =3/10
1/3 -1/x+1= 3/10
1/x+1= 1/3 -3/10
1/x+1 = 1/30
=> x+1= 30
x= 30-1
x= 29
Vậy...
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{11\cdot12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
c)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)