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BD
30 tháng 4 2019
b,
\(B=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
Ta thấy :
\(\frac{2018}{2019}>\frac{2018}{2019+2020}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020}\)
Từ đó , suy ra :
\(\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018+2019}{2019+2020}\)
Vậy...
#Louis
KN
30 tháng 4 2019
Ta có :
\(\frac{205}{321}< \frac{205}{315}\)
\(\frac{214}{315}>\frac{205}{315}\)
\(\Leftrightarrow\frac{205}{321}< \frac{214}{315}\)
M
0
19 tháng 11 2021
Bài 1 :
a) \(A=\frac{-1}{4.5}+\frac{-1}{5.6}-\frac{-1}{7.8}+\frac{-1}{9.10}\)
\(A=\frac{1}{4}\)\(-\left(-\frac{1}{5}\right)+...+\left(-\frac{1}{9}\right)-\left(-\frac{1}{10}\right)\)
\(A=\frac{1}{4}+\frac{1}{10}\)
\(A=\frac{3}{20}\)
19 tháng 11 2021
Bài 2:
a,17178585=1717:17178585:1717=15;13135151=1313:1015151:101=135115=51255<65255=1351⇒17178585<13135151a,17178585=1717:17178585:1717=15;13135151=1313:1015151:101=135115=51255<65255=1351⇒17178585<13135151
b,201201202202=201201:1001202202:1001=201202=201⋅1001001202⋅1001001=201201201202202202
DH
1
25 tháng 6 2015
a) \(\frac{214}{315}>\frac{205}{321}\)
b) \(\frac{2008}{2009}<\frac{10}{9}\)vì \(\frac{2008}{2009}<1;\frac{10}{9}>1\)
4 tháng 5 2017
\(A=\frac{100^{2007}+1}{100^{2008}+1}\Rightarrow100.A=\frac{100^{2008}+100}{100^{2008}+1}=\frac{100^{2008}+1+99}{100^{2008}+1}=1+\frac{99}{100^{2008}+1}\)
\(B=\frac{100^{2006}+1}{100^{2007}+1}\Rightarrow100.B=\frac{100^{2007}+100}{100^{2007}+1}=\frac{100^{2007}+1+99}{100^{2007}+1}=1+\frac{99}{100^{2007}+1}\)
Vì \(\frac{99}{100^{2007}+1}>\frac{99}{100^{2008}+1};1=1\Rightarrow1+\frac{99}{100^{2007}+1}>1+\frac{99}{100^{2008}+1}\)hay \(A>B\)
Vậy \(A>B\)
PT
5
MK
9 tháng 4 2017
1717/8585 = 17/85 = 1/5. 1313/5151=13/51. Mà 1/5 <13/51
Vay 1718 <1313/5151
9 tháng 4 2017
\(\frac{17}{85}vs\frac{13}{51}=\frac{1}{5}vs\frac{1}{3}\)
ta thấy 5>3
=>\(\frac{1717}{8585}< \frac{1313}{5151}\)
C
2
HT
18 tháng 8 2015
\(\frac{a}{b}=\frac{a\left(b+2015\right)}{b\left(b+2015\right)}=\frac{ab+2015a}{b\left(b+2015\right)}\)
\(\frac{a+2015}{b+2015}=\frac{b\left(a+2015\right)}{b\left(b+2015\right)}=\frac{ab+2015b}{b\left(b+2015\right)}\)
TH1: a = b
=> ab+2015a = ab+2015b
=> \(\frac{a}{b}=\frac{a+2015}{b+2015}\)
TH2: a > b
=> ab+2015a > ab+2015b
=> \(\frac{a}{b}>\frac{a+2015}{b+2015}\)
TH3: a < b
=> ab+2015a < ab+2015b
=> \(\frac{a}{b}<\frac{a+2015}{b+2015}\)
LC
18 tháng 8 2015
Ta có:\(\frac{a}{b}=\frac{a.\left(b+2015\right)}{b.\left(b+2015\right)}=\frac{ab+a.2015}{b.\left(b+2015\right)}\)
\(\frac{a+2015}{b+2015}=\frac{b.\left(a+2015\right)}{b.\left(b+2015\right)}=\frac{ab+b.2015}{b.\left(b+2015\right)}\)
Xét a>b=>a.2015>b.2015
=>\(\frac{ab+a.2015}{b.\left(b+2015\right)}>\frac{ab+b.2015}{b.\left(b+2015\right)}\)
=>\(\frac{a}{b}>\frac{a+2015}{b+2015}\)
Xét a=b=>a.2015=b.2015
=>\(\frac{ab+a.2015}{b.\left(b+2015\right)}=\frac{ab+b.2015}{b.\left(b+2015\right)}\)
=>\(\frac{a}{b}=\frac{a+2015}{b+2015}\)
Xét a<b=>a.2015<b.2015
=>\(\frac{ab+a.2015}{b.\left(b+2015\right)}<\frac{ab+b.2015}{b.\left(b+2015\right)}\)
=>\(\frac{a}{b}<\frac{a+2015}{b+2015}\)
Vậy \(\frac{a}{b}>\frac{a+2015}{b+2015}\)khi a>b
\(\frac{a}{b}=\frac{a+2015}{b+2015}\)khi a=b
\(\frac{a}{b}<\frac{a+2015}{b+2015}\)khi a<b
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