Đáp án:
`A=12,4`
Giải thích các bước giải:
Ta có :
`A=(\frac{2418}{1xx3}+\frac{2418}{3xx5}+\frac{2418}{5xx7}+...+\frac{2418}{193xx195}):97`
`→\frac{A}{1209}=(\frac{2}{1xx3}+\frac{2}{3xx5}+\frac{2}{5xx7}+...+\frac{2}{193xx195}):97`
`→\frac{A}{1209}=(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{193}-\frac{1}{195}):97`
`→\frac{A}{1209}=(\frac{1}{1}-\frac{1}{195}):97`
`→\frac{A}{1209}=\frac{194}{195}:97`
`→\frac{A}{1209}=\frac{2}{195}`
`→A=\frac{2}{195}xx1209`
`→A=\frac{62}{5}`
`→A=12,4`
Vậy `A=12,4`
