`color{#fc88d2}{} color{#fc88d2}{t} color{#fc88d2}{t} color{#fc88d2}{h} color{#fc88d2}{u} color{#fc88d2}{y} color{#6d79fc}{} color{#8892fc}{} color{#fc88d2}{}`

`A=1-2/(3xx5)-2/(5xx7)-2/(7xx9)-.......-2/(63xx65)`

`A=1-(2/(3xx5)+2/(5xx7)+2/(7xx9)-.......+2/(63xx65))`

`A=1-(1/3-1/5+1/5-1/7+1/7-1/9+.........+1/63-1/65)`

`A=1-(1/3-1/65)`

`A=1-(62/195)`

`A=133/195`

Vậy `A=133/195`