-Lời giải:
`A=1/21+1/22+1/23+.......+1/50`
`=(1/21+1/22+........+1/30)+(1/31+1/32+.......+1/40)+(1/41+1/42+......+1/50)`
Vì `1/21>1/30`
`1/22>1/30`
`......................`
`1/30=1/30`
`=>1/21+1/22+........+1/30>underbrace{1/30+1/30+.........+1/30}_{text{10 số}}`
`<=>1/21+1/22+........+1/30>10.1/30=1/3(1)`
Vì `1/31>1/40`
`1/32>1/40`
`.....................`
`1/40=1/40`
`=>1/31+1/32+.......+1/40>underbrace{1/40+1/40+.........+1/40}_{text{10 số}}`
`=>1/31+1/32+.......+1/40>10.1/40=1/4(2)`
Vì `1/41>1/50`
`1/42>/50`
`...................`
`1/50=1/50`
`=>1/41+1/42+......+1/50>underbrace{1/50+1/50+.........+1/50}_{text{10 số}}`
`<=>1/41+1/42+......+1/50>10.1/50=1/5(3)`
Cộng từng vế của `(1)(2)(3)` ta có:
`(1/21+1/22+........+1/30)+(1/31+1/32+.......+1/40)+(1/41+1/42+......+1/50)>1/3+1/4+1/5`
`<=>(1/21+1/22+........+1/30)+(1/31+1/32+.......+1/40)+(1/41+1/42+......+1/50)>47/60`
Mà `47/60>45/60=3/4`
`<=>(1/21+1/22+........+1/30)+(1/31+1/32+.......+1/40)+(1/41+1/42+......+1/50)>3/4`
Hay `A>B.`
Vậy `A>B`.
-Giải thích:Ta sử dụng tính chất `0<a<b=>1/a>1/b.`
