Question

Tính giá trị biểu thức
2) K= \(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+...+ \(\frac{1}{9900}\)

3) L= \(\frac{5}{1.4}\)+\(\frac{5}{4.7}\)+ .... + \(\frac{5}{100.103}\)

Answer 1

\(\left(2\right)K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(K=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(K=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\left(3\right)L=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{100\cdot103}\)

\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{510}{309}=\frac{170}{103}\)

Answer 2

Trả lời:

2,\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(K=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(K=\frac{1}{2}-\frac{1}{100}\)

\(K=\frac{49}{100}\)

3,\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(L=\frac{5}{3}\times\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{103}\right)\)

\(L=\frac{5}{3}\times\frac{102}{103}\)

\(L=\frac{170}{103}\)

Học tốt 

Answer 3

2)\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

3)\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

Answer 4

Tính giá trị biểu thức

2, K = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

K = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

K =\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

K = \(\frac{1}{2}-\frac{1}{100}\)

K = \(\frac{50}{100}-\frac{1}{100}\)

K = \(\frac{49}{100}\)

Answer 5

^{Thanks mọi người nha!} _{Mai mình đi học nốt buổi cuối phải nộp rồi!}

Answer 6

3, L = \(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

L = \(\frac{5}{3}.(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103})\)

L = \(\frac{5}{3}.(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

L = \(\frac{5}{3}.(1-\frac{1}{103})\)

L = \(\frac{5}{3}.\frac{102}{103}\)

L = \(\frac{170}{103}\)