các bạn giúp mình nha!!!
Ta có: 3A= 3/2.5+3/5.8+...+3/92.95+3/95.98
= 1/2-1/5+1/5-1/8+....+1/95-1/98
1/2-1/98=24/49
=> A=(24/49);3=8/49
\(A=\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{92\times95}+\frac{1}{95\times98}\)
\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{92\times95}+\frac{3}{95\times98}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{1}{3}\times\frac{24}{49}=\frac{8}{49}\)
Chúc bạn học tốt
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{24}{49}\)
\(A=\frac{4}{45}\)
các bạn giúp mình nha!!!
