\(P=\dfrac{12}{1\cdot4\cdot7}+\dfrac{12}{4\cdot7\cdot10}+\dfrac{12}{7\cdot10\cdot13}+...+\dfrac{12}{54\cdot57\cdot60}\)

\(P=\dfrac{12}{6}\left(\dfrac{1}{1\cdot4}-\dfrac{1}{4\cdot7}+\dfrac{1}{4\cdot7}-\dfrac{1}{7\cdot10}+...+\dfrac{1}{54\cdot57}-\dfrac{1}{57\cdot60}\right)\)

\(P=2\left(\dfrac{1}{1\cdot4}-\dfrac{1}{57\cdot60}\right)\)

\(P=\dfrac{2}{4}-\dfrac{2}{57\cdot60}=\dfrac{1}{2}-\dfrac{1}{57\cdot30}\)

\(\Rightarrow P< \dfrac{1}{2}\)

=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

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)

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 < \(\frac{1}{2}\)

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

Ta có \(A=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-...+\frac{1}{16.19}-\frac{1}{19.22}\)

\(=\frac{1}{4}-\frac{1}{418}=\frac{207}{836}\)

\(A=\frac{6}{1\cdot4\cdot7}+\frac{6}{4\cdot7\cdot10}+\frac{6}{7\cdot10\cdot13}+...+\frac{6}{16\cdot19\cdot22}\)

\(A=\frac{1}{1\cdot4}-\frac{1}{4\cdot7}+\frac{1}{4\cdot7}-\frac{1}{7\cdot10}+...+\frac{1}{16\cdot19}-\frac{1}{19\cdot22}\)

\(A=\frac{1}{4}-\frac{1}{19\cdot22}=\frac{207}{836}\)

S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)

= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)

S=P nhé