Question

Tính : A = \(\frac{1}{3}\times\frac{1}{7}+\frac{1}{7}\times\frac{1}{11}+\frac{1}{11}\times\frac{1}{15}+...+\frac{1}{95}\times\frac{1}{99}\)

Answer 1

\(4A=\frac{4}{3.7}+...+\frac{4}{95.99}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(4A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(\Rightarrow A=\frac{8}{99}\)

Answer 2

\(A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}\)

\(A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\frac{32}{99}\)

\(A=\frac{8}{99}\)

Answer 3

\(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}.\)

    \(=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)

\(4A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(4A=\frac{1}{3}-\frac{1}{99}\)\(=\frac{33-1}{99}\)\(=\frac{32}{99}\)

\(\Rightarrow A=\frac{32}{99}:4=\frac{32}{99}.\frac{1}{4}=\frac{8}{99}\)

Answer 4

$A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}$

$A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)$

$A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)$

$A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)$

$A=\frac{1}{4}.\frac{32}{99}$

$A=\frac{8}{99}$

Answer 5

A=\(\frac{1}{3}\)X\(\frac{1}{7}+\frac{1}{7}X\frac{1}{11}+\frac{1}{11}X\frac{1}{15}+.....+\frac{1}{95}+\frac{1}{99}\)

A=\(\frac{1}{3X7}+\frac{1}{7X11}+\frac{1}{11X15}+...+\frac{1}{95X99}\)

Suy ra 4A=\(4\times(\frac{1}{3\times7}+\frac{1}{7\times11}+\frac{1}{11\times15}+...+\frac{1}{95\times99})\)

4A=\(\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{95\times99}\)

4A=\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)

4A=\(\frac{1}{3}-\frac{1}{99}\)

4A=\(\frac{32}{99}\)

\(\Rightarrow A=\frac{32}{99}:4=\frac{8}{99}\)

Nhớ tích cho mk nha

Answer 6

A= \(\frac{1}{3}\)x \(\frac{1}{7}\)+ \(\frac{1}{7}\)x \(\frac{1}{11}\)+ \(\frac{1}{11}\)x \(\frac{1}{15}\)+ ... + \(\frac{1}{95}\)x \(\frac{1}{99}\)

= \(\frac{1}{3}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{11}\)+ \(\frac{1}{11}\)- \(\frac{1}{15}\)+ ... + \(\frac{1}{95}\)- \(\frac{1}{99}\)

= \(\frac{1}{99}\)- \(\frac{1}{3}\)

= \(\frac{32}{99}\)

Answer 7

\(A=\frac{1}{3}\text{ x }\frac{1}{7}+\frac{1}{7}\text{ x }\frac{1}{11}+\frac{1}{11}\text{ x }\frac{1}{15}+...+\frac{1}{95}\text{ x }\frac{1}{99}\)

\(A=\frac{1}{3\text{ x }7}+\frac{1}{7\text{ x }11}+\frac{1}{11\text{ x }15}+...+\frac{1}{95\text{ x }99}\)

\(4A=\frac{4}{3\text{ x }7}+\frac{4}{7\text{ x }11}+\frac{4}{11\text{ x }15}+...+\frac{4}{95\text{ x }99}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)

\(4A=\frac{1}{3}-\frac{1}{99}\)

\(4A=\frac{32}{99}\)

\(A=\frac{32}{99}\text{ : }4\)

\(A=\frac{8}{99}\)

Answer 8

  A = \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)

 4A =  \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

 4A  = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

 4A = \(\frac{1}{3}-\frac{1}{99}\)= \(\frac{32}{99}\)

  A = \(\frac{32}{99}:4=\frac{8}{99}\)

 ~ HỌC TỐT ~