Bài 2 So sánh A=2/60.63+2/63.66+⋯+2/117.120+2/2002 B=5/40.44+5/44.48+⋯+5/76.80+5/2003 13/07/2021 Bởi Aubrey Bài 2 So sánh A=2/60.63+2/63.66+⋯+2/117.120+2/2002 B=5/40.44+5/44.48+⋯+5/76.80+5/2003
Đáp án: `A<B` Giải thích các bước giải: `A=2/60.63+2/63.66+…+2/117.120+2/2002` `=>A/2=1/60.63+1/63.66+…+1/117.120+1/2002` `=>(3A)/2=3/60.63+3/63.66+…+3/117.120+3/2002` `=>(3A)/2=1/60-1/63+1/63-1/66+…+1/117-1/120+3/2002` `=>(3A)/2=1/60-1/120+3/2002` `=>(3A)/2=1/120+3/2002` `=>(3A)/2=1181/120120` `=>3A=1181/120120*2` `=>3A=1181/60060` `=>A=1181/60060÷3` `=>A=1181/180180` `B=5/40.44+5/44.48+…+5/76.80+5/2003` `=>B/5=1/40.44+1/44.48+…+1/76.80+1/2003` `=>(4B)/5=4/40.44+4/44.48+…+4/76.80+4/2003` `=>(4B)/5=1/40-1/44+1/44-1/48+…+1/76-1/80+4/2003` `=>(4B)/5=1/40-1/80+4/2003` `=>(4B)/5=1/80+4/2003` `=>(4B)/5=1683/160240` `=>4B=1683/160240*5` `=>4B=1683/32048` `=>B=1683/32048÷4` `=>B=1683/128192` `=>A=37848688/5774408640<75810735/5774408640=B` `=>A<B` Vậy `A<B`. Bình luận
Đáp án: B > A Giải thích các bước giải: Ta có : A = $\frac{2}{60.63}$ +$\frac{2}{63.66}$ +…+$\frac{2}{117.120}$ + $\frac{2}{2002}$ = ($\frac{1}{60}$ -$\frac{1}{63}$)+($\frac{1}{63}$ -$\frac{1}{66}$)+…+($\frac{1}{117}$ -$\frac{1}{120}$) + $\frac{2}{2002}$ = $\frac{1}{60}$ -$\frac{1}{63}$ + $\frac{1}{63}$ -$\frac{1}{66}$ + … + $\frac{1}{117}$ -$\frac{1}{120}$ + $\frac{2}{2002}$ = $\frac{1}{60}$ – $\frac{1}{120}$ + $\frac{2}{2002}$ = $\frac{1}{120}$ + $\frac{2}{2002}$ . B = $\frac{5}{40.44}$ + $\frac{5}{44.48}$ +…+ $\frac{5}{76.80}$ + $\frac{5}{2003}$ ⇒ $\frac{4}{5}$B = $\frac{4}{40.44}$ + $\frac{4}{44.48}$ +…+ $\frac{4}{76.80}$ + $\frac{4}{2003}$ Tương tự = $\frac{1}{40}$ – $\frac{1}{80}$ + $\frac{4}{2003}$ = $\frac{1}{80}$ + $\frac{4}{2003}$ ⇒ B = $\frac{5}{320}$ + $\frac{5}{2003}$ = $\frac{1}{64}$ + $\frac{5}{2003}$ Khi đó : B > A. Bạn thử xem lại đề bài xem có đề có sai sót gì không. Bình luận
Đáp án:
`A<B`
Giải thích các bước giải:
`A=2/60.63+2/63.66+…+2/117.120+2/2002`
`=>A/2=1/60.63+1/63.66+…+1/117.120+1/2002`
`=>(3A)/2=3/60.63+3/63.66+…+3/117.120+3/2002`
`=>(3A)/2=1/60-1/63+1/63-1/66+…+1/117-1/120+3/2002`
`=>(3A)/2=1/60-1/120+3/2002`
`=>(3A)/2=1/120+3/2002`
`=>(3A)/2=1181/120120`
`=>3A=1181/120120*2`
`=>3A=1181/60060`
`=>A=1181/60060÷3`
`=>A=1181/180180`
`B=5/40.44+5/44.48+…+5/76.80+5/2003`
`=>B/5=1/40.44+1/44.48+…+1/76.80+1/2003`
`=>(4B)/5=4/40.44+4/44.48+…+4/76.80+4/2003`
`=>(4B)/5=1/40-1/44+1/44-1/48+…+1/76-1/80+4/2003`
`=>(4B)/5=1/40-1/80+4/2003`
`=>(4B)/5=1/80+4/2003`
`=>(4B)/5=1683/160240`
`=>4B=1683/160240*5`
`=>4B=1683/32048`
`=>B=1683/32048÷4`
`=>B=1683/128192`
`=>A=37848688/5774408640<75810735/5774408640=B`
`=>A<B`
Vậy `A<B`.
Đáp án:
B > A
Giải thích các bước giải:
Ta có : A = $\frac{2}{60.63}$ +$\frac{2}{63.66}$ +…+$\frac{2}{117.120}$ + $\frac{2}{2002}$
= ($\frac{1}{60}$ -$\frac{1}{63}$)+($\frac{1}{63}$ -$\frac{1}{66}$)+…+($\frac{1}{117}$ -$\frac{1}{120}$) + $\frac{2}{2002}$
= $\frac{1}{60}$ -$\frac{1}{63}$ + $\frac{1}{63}$ -$\frac{1}{66}$ + … + $\frac{1}{117}$ -$\frac{1}{120}$ + $\frac{2}{2002}$
= $\frac{1}{60}$ – $\frac{1}{120}$ + $\frac{2}{2002}$
= $\frac{1}{120}$ + $\frac{2}{2002}$ .
B = $\frac{5}{40.44}$ + $\frac{5}{44.48}$ +…+ $\frac{5}{76.80}$ + $\frac{5}{2003}$
⇒ $\frac{4}{5}$B = $\frac{4}{40.44}$ + $\frac{4}{44.48}$ +…+ $\frac{4}{76.80}$ + $\frac{4}{2003}$
Tương tự = $\frac{1}{40}$ – $\frac{1}{80}$ + $\frac{4}{2003}$
= $\frac{1}{80}$ + $\frac{4}{2003}$
⇒ B = $\frac{5}{320}$ + $\frac{5}{2003}$
= $\frac{1}{64}$ + $\frac{5}{2003}$
Khi đó : B > A.
Bạn thử xem lại đề bài xem có đề có sai sót gì không.