so sánh các biểu thức sau: A = 1/21 + 1/22 + 1/23 + … + 1/50 và B = 3/4

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1. -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.`

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2. `A=1/21+1/22+1/23+…+1/50`

   `⇒A=(1/21+1/22+…+1/30)+(1/31+1/32+…+1/40)+(1/31+1/32+…+1/40)`

   `⇒A>(\underbrace{1/30+1/30+…+1/30}_{\text{10 số hạng}})+(\underbrace{1/40+1/40+…+1/40}_{\text{10 số hạng}})+(\underbrace{1/50+1/50+…+1/50}_{\text{10 số hạng}})`

   `⇒A>10. 1/30+10. 1/40+10. 1/50`

   `⇒A>10/30+10/40+10/50`

   `⇒A>1/3+1/4+1/5`

   `⇒A>{20+15+12}/60`

   `⇒A>47/60>45/60=3/4`

   `⇒A>B`

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