Tính:
\(\frac{1}{1.2}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{5.6}\)+ ......+\(\frac{1}{9.10}\)
\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{9.10}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)
\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)
\(=-\frac{9}{10}+\frac{1}{90}\)
= ...
bn tự tính nha!
Tính theo cách hợp lí:
M = \(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(M=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{9.10}\) và \(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\). Tính A : B
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{9.10}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\)
\(\Rightarrow A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
\(\Rightarrow A=\left(\frac{1}{6}+\frac{1}{10}\right)+\left(\frac{1}{7}+\frac{1}{9}\right)+\frac{1}{8}\)
\(\Rightarrow A=\left(\frac{10}{6.10}+\frac{6}{6.10}\right)+\left(\frac{9}{7.9}+\frac{7}{7.9}\right)+\frac{8}{8.8}\)
\(\Rightarrow A=\frac{16}{6.10}+\frac{16}{7.9}+\frac{8}{8.8}\)
\(\Rightarrow A=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)\)
Ta lại có:
\(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\)
\(\Rightarrow B=\left(\frac{1}{6.10}+\frac{1}{6.10}\right)+\left(\frac{1}{7.9}+\frac{1}{7.9}\right)+\frac{1}{8.8}\)
\(\Rightarrow B=\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\)
Vậy :
\(A:B=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right):\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)=8\)
Vậy \(A:B=8\)
Chứng minh rằng:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Đây là tính chứ chứng minh cái gì ?
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Lập luận: 1/1.2 = 1/1 - 1/2 ; 1/2.3 = 1/2 - 1/3 ; 1/3.4 = 1/3 - 1/4 ; làm tương tự với các số kia.
Ta có: 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10
= 1 - 1/10
= 9/10
Tính A : B biết:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\) và \(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\)
A = 1/1.2 + 1/3.4 + 1/5.6 + 1/7.8 + 1/9.10
A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10
A = ( 1 + 1/3 + 1/5 + 1/7 + 1/9) - ( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)
A = ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - 2.( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)
A = ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - ( 1 + 1/2 + 1/3 + 1/4 + 1/5)
A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
B = 1/6.10 + 1/7.9 + 1/8.8 + 1/9.7 + 1/10.6
16B = 16/6.10 + 16/7.9 + 16/8.8 + 16/9.7 + 16/10.6
16B = 1/6 + 1/10 + 1/7 + 1/9 + 1/8 + 1/8 + 1/9 + 1/7 + 1/10 + 1/6
16B = 2.( 1/6 + 1/7 + 1/8 + 1/9 + 1/10)
8B = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
Ta có A = 8B
=> A : B = 8
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}\)
\(A=\frac{2}{5}\)
\(\text{A = 1/1.2 + 1/3.4 + 1/5.6 + 1/7.8 + 1/9.10 }\)
\(\text{= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10
}\)\(\text{= ( 1 + 1/3 + 1/5 + 1/7 + 1/9) - ( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)}\)
= ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - 2.( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)
= ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - ( 1 + 1/2 + 1/3 + 1/4 + 1/5)
= 1/6 + 1/7 + 1/8 + 1/9 + 1/10
B = 1/6.10 + 1/7.9 + 1/8.8 + 1/9.7 + 1/10.6 16
= 16/6.10 + 16/7.9 + 16/8.8 + 16/9.7 + 16/10.6 16
= 1/6 + 1/10 + 1/7 + 1/9 + 1/8 + 1/8 + 1/9 + 1/7 + 1/10 + 1/6
= 2.( 1/6 + 1/7 + 1/8 + 1/9 + 1/10)
= 1/6 + 1/7 + 1/8 + 1/9 + 1/10
Ta có A = 8B => A : B = 8
Tính Tỉ Số A và B biết :
A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{9.10}\)
B = \(\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\)
a/ 1/1.3 + 1/3.4+ .... + 1/9.10
= 1–1/3+1/3–1/4+1/4–1/5+...+1/8–1/9+1/9–1/10
=1–1/10
=10/10–1/10
=9/10
Chứng minh rằng:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
b)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< 1-\frac{1}{2.3}\)
Cần gấp, ai nhanh mik tick nha
Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
Tính nhanh
\(B=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)
\(B=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{3}-\frac{1}{10}\)
\(B=\frac{7}{30}\)
Tính
\(\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{10}\)
\(=\frac{1}{10}\)
(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-......+1/9-1/10)
1-1/10=9/10
nhớ cho mk