Tính hợp lí
6/1.4.7+6/4.7.10+....+6/54.57.60
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}=A\)
\(\frac{A}{2}=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(\frac{A}{2}=\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{60-54}{54.57.60}\)
\(\frac{A}{2}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{1.4}-\frac{1}{57.60}\)
\(A=\frac{1}{2}-\frac{1}{30.57}< \frac{1}{2}\)
\(A=\frac{1}{1.4.7}+\frac{1}{4.7.10}+...+\frac{1}{54.57.60}\)
\(\Rightarrow6A=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.47}-\frac{1}{57.60}\)
\(=\frac{1}{4}-\frac{1}{3420}=\frac{855}{3420}-\frac{1}{3420}=\frac{427}{1710}\)
\(\Rightarrow A=\frac{427}{1710}:6=\frac{427}{1710}.\frac{1}{6}=\frac{427}{10260}\)
Nhận thấy:
\(\frac{6}{1.4.7}=\frac{1}{1.4}-\frac{1}{4.7}\)
...............
\(\frac{6}{54.57.60}=\frac{1}{54.57}-\frac{1}{57.60}\)
=> ta phải nhân A vói 6
=> 6A =
\(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{4}-\frac{1}{57.60}=\frac{427}{1710}\)
=> A = 427/1710 : 6 =427/10260
Gọi biểu thức là A, ta có:
A = \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}=2\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+\frac{6}{7.10.13}+...+\frac{6}{54.57.60}\right)\)
A = \(2\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-\frac{1}{10.13}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
A = \(2\left(\frac{1}{1.4}-\frac{1}{57.60}\right)=2\left(\frac{427}{1710}\right)=\frac{427}{855}< \frac{427}{854}=\frac{1}{2}\)
Vậy A < \(\frac{1}{2}\)(điều cần chứng minh)
P = 2*[ 6/(1*4*7) + 6/(4*7*10) + ... + 6/(54*57*60) ]
= 2*[ 1/(1*4) - 1/(4*7) + 1/(4*7) - 1/(7*10) + ... + 1/(54*57) -1/(57*60) ]
= 2*[ 1/(1*4) - 1/(57*60) ]
= 2* (427/1710)
= 427/855 <1/2
S = 1+ 1/2^2 + 1/3^2 +... + 1/100^2
1/2^2 < 1/(1*2)
1/3^2 < 1/(2*3)
...
1/100^2 < 1/(99*100)
==> 1/2^2 +1/3^2 +.., +1/100^2 < 1/(1*2) + 1/(2*3) + ... + 1/(99*100) = 1 -1/2 +1/2 - 1/3 +1/3 -1/4 +... - 1/100
=1 - 1/100 <1
==> 1/2^2 + 1/3^2 +... + 1/100^2 < 1
==> 1 + 1/2^2 + 1/3^2 +... +1/100^2 <2
Ta có \(A=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-...+\frac{1}{16.19}-\frac{1}{19.22}\)
\(=\frac{1}{4}-\frac{1}{418}=\frac{207}{836}\)
\(A=\frac{6}{1\cdot4\cdot7}+\frac{6}{4\cdot7\cdot10}+\frac{6}{7\cdot10\cdot13}+...+\frac{6}{16\cdot19\cdot22}\)
\(A=\frac{1}{1\cdot4}-\frac{1}{4\cdot7}+\frac{1}{4\cdot7}-\frac{1}{7\cdot10}+...+\frac{1}{16\cdot19}-\frac{1}{19\cdot22}\)
\(A=\frac{1}{4}-\frac{1}{19\cdot22}=\frac{207}{836}\)
\(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(=3.\left(\frac{1}{1.4}-\frac{1}{4.7}\right)+3.\left(\frac{1}{4.7}-\frac{1}{7.10}\right)+...+3.\left(\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(=3\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...-\frac{1}{54.57}+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(=3\left(\frac{1}{1.4}-\frac{1}{57.60}\right)\)
\(=3\left(\frac{1}{4}-\frac{1}{3420}\right)\)
\(=3\left(\frac{855}{3420}-\frac{1}{3420}\right)\)
\(=3.\frac{427}{1710}\)
\(=\frac{427}{570}\)