Bài 1:So sánh
A= 2/60×63 +2/63×66 + ….+ 2/117×120 + 2/120×123 với B= 5/40×44 + 5/44×48 + …..+5/76×80 + 5/2006
Bài 2:Tính
B= 3 + 3/1+2 +3/1+2+3 + 3/1+2+3+…….+3/1+2+3+4+………+100
Bài 1:So sánh A= 2/60×63 +2/63×66 + ….+ 2/117×120 + 2/120×123 với B= 5/40×44 + 5/44×48 + …..+5/76×80 + 5/2006 Bài 2:Tính B= 3 + 3/1+2 +3/1+2+3 + 3
By aihong
Đáp án:
Giải thích các bước giải:
`1`
`A=2/60.63+2/63.66+….+2/117.120+2/120.123`
`=>A/2=1/60.63+1/63.66+…+1/117.120+1/120.123`
`=>(3A)/2=3/60.63+3/63.66+…+3/117.120+3/120.123`
`=>(3A)/2=1/60-1/63+1/63-1/66+…+1/117-1/120+1/120-1/123`
`=>(3A)/2=1/60-1/123`
`=>(3A)/2=7/820`
`=>3A=7/820*2`
`=>3A=7/410`
`=>A=7/410÷3`
`=>A=7/1230`
`B=5/40.44+5/44.48+…+5/76.80+5/2006`
`=>B/5=1/40.44+1/44.48+…+1/76.80+1/2006`
`=>(4B)/5=4/40.44+4/44.48+…+4/76.80+4/2006`
`=>(4B)/5=1/40-1/44+1/44-1/48+…+1/76-1/80+4/2006`
`=>(4B)/5=1/40-1/80+4/2006`
`=>(4B)/5=1/80+4/2006`
`=>(4B)/5=1163/80240`
`=>4B=1163/80240*5`
`=>4B=1163/16048`
`=>B=1163/16048÷4`
`=>B=1163/64192`
`=>A=224672/39478080<715245/39478080=B`
`=>A<B`
Vậy `A<B`.
`2`
`B=3+3/(1+2)+3/(1+2+3)+3/(1+2+3+4)+…+3/(1+2+3+4+…+100)`
`=>B=3+3/3+3/6+3/10+…+`$\dfrac{3}{\dfrac{(100+1).100}{2}}$ `=>B=3+3/3+3/6+3/10+…+`$\dfrac{3}{\dfrac{101.100}{2}}$
`=>B/2=3/2+3/6+3/12+3/20+…+3/(100.101)`
`=>B/2=3/1.2+3/2.3+3/3.4+3/4.5+…+3/100.101`
`=>B/6=1/1.2+1/2.3+1/3.4+1/4.5+…+1/100.101`
`=>B/6=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/100-1/101`
`=>B/6=1-1/101`
`=>B/6=100/101`
`=>B=100/101*6`
`=>B=600/101`
Vậy `B=600/101`.
Đáp án : Bài 1: $B>A$
Giải thích các bước giải:
Bài 1:
Ta có :
$A=\dfrac{2}{60\times 63}+\dfrac{2}{63\times 66}+…+\dfrac{2}{117\times 120}+\dfrac{2}{120\times 123}$
$\rightarrow A=2(\dfrac{1}{60\times 63}+\dfrac{1}{63\times 66}+…+\dfrac{1}{117\times 120}+\dfrac{1}{120\times 123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{3}{60\times 63}+\dfrac{3}{63\times 66}+…+\dfrac{3}{117\times 120}+\dfrac{3}{120\times 123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{63-60}{60\times 63}+\dfrac{66-63}{63\times 66}+…+\dfrac{120-117}{117\times 120}+\dfrac{123-120}{120\times 123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+..+\dfrac{1}{117}-\dfrac{120}+\dfrac{1}{120}-\dfrac{1}{123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{1}{60}-\dfrac{1}{123})$
Lại có :
$B=\dfrac{5}{40\times 44}+\dfrac{5}{44\times 48}+..+\dfrac{5}{76\times 80}+\dfrac{5}{2006}$
$\rightarrow B=5(\dfrac{1}{40\times 44}+\dfrac{1}{44\times 48}+..+\dfrac{1}{76\times 80}+\dfrac{1}{2006})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{4}{40\times 44}+\dfrac{4}{44\times 48}+..+\dfrac{4}{76\times 80}+\dfrac{4}{2006})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{44-40}{40\times 44}+\dfrac{48-44}{44\times 48}+..+\dfrac{80-76}{76\times 80}+\dfrac{4}{2006})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+..+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{2}{80}-\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B>1>\dfrac{1}{2}.\dfrac{1}{60}>A$