Chứng minh rằng:
\(\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}< \frac{1}{2}\)
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)
CHỨNG MINH RẰNG:\(\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}< \frac{1}{2}\)
Đặt \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}=A\)
\(\frac{A}{2}=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(\frac{A}{2}=\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{60-54}{54.57.60}\)
\(\frac{A}{2}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{1.4}-\frac{1}{57.60}\)
\(A=\frac{1}{2}-\frac{1}{30.57}< \frac{1}{2}\)
chứng minh rằng : P = \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54+57+60}< \frac{1}{2}\)
Câu hỏi của thục hà - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo nhé!
Đề sai hả
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{57.60}< \frac{1}{4}\)
\(\Rightarrow P< \frac{1}{4}.2=\frac{1}{2}\)
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}\)
\(=2\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\right)\)
\(=2\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{3420}\right)\)
\(=2\left(\frac{855-1}{3420}\right)\)
\(=2.\frac{427}{1710}=\frac{427}{855}\)
Mà \(\frac{1}{2}=\frac{427}{854}\)
=> \(\frac{427}{855}< \frac{427}{854}\)=> P < \(\frac{1}{2}\)
Chứng Minh
\(\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.12}+.......+\frac{12}{54.57.60}<\frac{1}{2}\)
Tính nhanh:
P=\(\frac{12}{1.4.7}\)+ \(\frac{12}{4.7.10}\)+\(\frac{12}{7.10.13}\)+....+ \(\frac{12}{54.57.60}\)
Giúp e vs e đang cần gấp!!
Đúng em tick cho
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}\)
\(P=4.\left(\frac{3}{1.4.7}+\frac{3}{4.7.10}+\frac{3}{7.10.13}+...+\frac{3}{54.57.60}\right)\)
\(P=4\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(P=4.\left(\frac{1}{4}-\frac{1}{3420}\right)\)
\(P=4.\frac{427}{1710}\)
\(P=\frac{854}{855}\)
Tính nhanh:
P=\(\frac{12}{1.4.7}\)+\(\frac{12}{4.7.10}\)+\(\frac{12}{7.10.13}\)+......+\(\frac{12}{54.57.60}\)
Mọi người giúp em nhé.
nhớ nhiều nhé
duyệt đi mà nhanh lên sốt ruột quá
Cho \(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+.....+\frac{12}{55.58.61}SosánhVớiP\frac{1}{2}\)
Tính nhanh:
P=12/1.4.7 + 12/4.7.10 + 12/7.10.13 + ... + 12/54.57.60
=2.(6/1.4.7 + 6/4.7.10 + 6/7.10.13 + ... + 6/54.57.60)
=2.(1/1.4-1/4.7+1/4.7-1/7.10+1/7.10-1/10.13+...+1/54.57-1/57.60)
=2(1.4-1/57.60)
TỰ TÍNH
1.tính
B= 12/ 1.4.7 + 12/ 4.7.10 + 12/ 7.10.13 +....+ 12/54.57.60
2.chứng minh
a) A= 1+ 1/2^2 + 1/3^2 +...+1/100^2 < 2
b) B= 1/5 + 1/13 +1/25 +...+1/n^2+(n+1)^2< 1/2
1
B= 12/1.4.7 + 12/4.7.10 + 12/7.10.13 + ... + 12/54.57.60
=> 1/2B= 6/1.4.7 + 6/4.7.10 + 6/7.10.13 + ... + 6/54.57.60
=> 1/2B = 1/1.4 - 1/4.7 +1/4.7 - 1/7.10 +1/7.10 - 1/10.13 + ... + 1/54.57 - 1/57.60
=> 1/2B =1/1.4 - 1/57.60
=> 1/2B = 1/4 - 1/3420
=> 1/2B = 427/1710
=> B = 427/1710 . 2
=> B = 427/855
2
A= 1+ 1/22 + 1/32 +...+1/1002
=1+ 1/2.2 + 1/3.3 +...+ 1/100.100
=> A< 1+ 1/1.2 + 1/2.3 +...+ 1/99.100
= 1+ 1 - 1/2 +1/2 - 1/3 +...+1/99 - 1/100
= 2- 1/100 < 2
Vậy A < 2
