\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\) 

\(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}A=1-\frac{1}{10}\)

\(\frac{1}{2}A=\frac{9}{10}\)

\(A=\frac{9}{10}:\frac{1}{2}\)

\(A=\frac{18}{10}=\frac{9}{5}\)

sao thế tự dưng lấy 1/12 ở đâu thế 1/10 cơ mà

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

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

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

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

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

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 < \(\frac{1}{2}\)

\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}\)

\(P=4.\left(\frac{3}{1.4.7}+\frac{3}{4.7.10}+\frac{3}{7.10.13}+...+\frac{3}{54.57.60}\right)\)

\(P=4\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)

\(P=4.\left(\frac{1}{4}-\frac{1}{3420}\right)\)

\(P=4.\frac{427}{1710}\)

\(P=\frac{854}{855}\)

nhớ nhiều nhé

duyệt đi mà nhanh lên sốt ruột quá

\(A=\frac{1}{1.4.7}+\frac{1}{4.7.10}+...+\frac{1}{54.57.60}\)

\(\Rightarrow6A=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)

\(=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.47}-\frac{1}{57.60}\)

\(=\frac{1}{4}-\frac{1}{3420}=\frac{855}{3420}-\frac{1}{3420}=\frac{427}{1710}\)

\(\Rightarrow A=\frac{427}{1710}:6=\frac{427}{1710}.\frac{1}{6}=\frac{427}{10260}\)

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

Đặt \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}=A\)

\(\frac{A}{2}=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)

\(\frac{A}{2}=\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{60-54}{54.57.60}\)

\(\frac{A}{2}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{1.4}-\frac{1}{57.60}\)

\(A=\frac{1}{2}-\frac{1}{30.57}< \frac{1}{2}\)