Question

Bài 3 Tính giá trị biểu thức\(\left(1_{ },5\right).\frac{-2}{3}+\left(2,5-\frac{3}{4}\right):1\frac{3}{4}\)

B=\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

 

Answer 1

\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\frac{102}{103}\)

\(B=\frac{34}{103}\)

Answer 2

Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)

\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\frac{102}{103}\)

\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)

Answer 3

Thằng an mất dạy :))) 
#Lê_Linh

Answer 4

B = 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/100.103

= 1/3 ( 3/1.4 + 3/4.7 + ... + 3/100.103 )

= 1/3 ( 1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103 )

= 1/3 ( 1 - 1/103 )

= 1/3 . 102/103

= 34/103