Đáp án:
Giải thích các bước giải:
`text(B)=1/3+(1/16+1/19+1/21)+(1/61+1/72+1/83+1/94)`
`=>1/16<1/15;1/19<1/15;1/21<1/15`
`=>1/16+1/19+1/21<1/15+1/15+1/15` (Có `3` số hạng)
`=>1/16+1/19+1/21<1/5(1)`
`=>1/61<1/60;1/72<1/60;1/83<1/60;1/94<1/60`
`=>1/61+1/72+1/83+1/94<1/60+1/60+1/60+1/60` (Có `4` số hạng)
`=>1/61+1/72+1/83+1/94<1/15(2)`
Từ `(1)` và `(2)` ta có:
`=>1/16+1/19+1/21+1/61+1/72+1/83+1/94<1/5+1/15`
`=>1/3+1/16+1/19+1/21+1/61+1/72+1/83+1/94<1/3+1/5+1/15`
`=>1/3+1/16+1/19+1/21+1/61+1/72+1/83+1/94<8/15+1/15`
`=>1/3+1/16+1/19+1/21+1/61+1/72+1/83+1/94<3/5`
Vậy `text(B)<3/5`.
