Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: A= \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{90}\)
\(A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}\right)\)
A= B+C
Ta có: \(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\)
\(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>30.\dfrac{1}{60}=\dfrac{1}{2}\) (1)
Lại có: \(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>\dfrac{1}{90}+\dfrac{1}{90}+...+\dfrac{1}{90}\)
\(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>30.\dfrac{1}{90}=\dfrac{1}{3}\) (2)
Từ (1) và (2) => \(A>\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
Vậy \(A>\dfrac{5}{6}\)
Ta có: \(\frac{1}{31}+\frac{1}{32}+.....+\frac{1}{90}=\frac{1}{\frac{\left(90-31+1\right).\left(90+31\right)}{2}}=\frac{1}{3630}.\)
Mà \(\frac{5}{6}=\frac{5.605}{6.605}=\frac{3025}{3630}\)
Vì \(\frac{3025}{3630}>\frac{1}{3630}\)
Nên \(\frac{1}{31}+\frac{1}{32}+.....+\frac{1}{90}< \frac{5}{6}\)
A = 1/31 + 1/32 + 1/33 + ... + 1/89 + 1/90 ..... 5/6
A = 5/6 = 1/2 + 1/3
Ta đặt : B = 1/31 + 1/32 + 1/33 + ... + 1/60 ﴾ 30 phân số ﴿
C = 1/61 + 1/62 + 1/63 + .... + 1/90 ﴾ 30 phân số ﴿
Ta có : B = 1/31 + 1/32 + 1/33 + ... + 1/60 > 1/60 + 1/60 + 1/60 + ... + 1/60 = 30 x 1/60 = 1/2
C = 1/61 + 1/62 + 1/63 + ... + 190 > 1/90 + 1/90 + 1/90 + .... + 1/90 = 30 x 1/90 = 1/3
Vì A = B + C > 1/2 + 1/3 = 5/6 nên 1/31 + 1/32 + 1/33 + .. + 1/89 + 1/90 > 5/6
\(M=\dfrac{1}{31}+\dfrac{1}{32}+.................+\dfrac{1}{89}+\dfrac{1}{90}\)
\(\Leftrightarrow M=\left(\dfrac{1}{31}+\dfrac{1}{32}+.........+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+........+\dfrac{1}{90}\right)\)
Đặt :
\(A=\dfrac{1}{31}+\dfrac{1}{32}+.......+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+.......+\dfrac{1}{60}=\dfrac{1}{60}.30=\dfrac{1}{2}\)
\(B=\dfrac{1}{61}+\dfrac{1}{62}+.......+\dfrac{1}{90}>\dfrac{1}{90}+\dfrac{1}{90}+......+\dfrac{1}{90}=\dfrac{1}{90}.30=\dfrac{1}{3}\)
\(\Leftrightarrow M=\dfrac{1}{31}+\dfrac{1}{32}+.........+\dfrac{1}{89}+\dfrac{1}{90}=A+B< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow M< \dfrac{5}{6}\rightarrowđpcm\)
Ta có:\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{89}+\frac{1}{90}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{90}\right)\)
\(>\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\right)\)
có 30 số hạng 1/60 có 30 số hạng 1/90
\(=\frac{30}{60}+\frac{30}{90}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
=> \(\frac{1}{31}+...+\frac{1}{90}>\frac{5}{6}\)
đây là cách ngắn gọn chỉ dành cho hs khá giỏi nha
\(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\)
\(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>\dfrac{1}{90}+\dfrac{1}{90}+...+\dfrac{1}{90}=\dfrac{30}{90}=\dfrac{1}{3}\)
Do đó: \(B+C>\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(đpcm)