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\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)và \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
Xét \(A=\frac{2019^{2020}+1}{2019^{2021}+1}\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)
Xét \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}=1+\frac{2018}{2019^{2019}+1}\)
Vì \(1+\frac{2018}{2019^{2021}+1}< 1+\frac{2018}{2019^{2019}+1}\Rightarrow\frac{2019^{2020}+1}{2019^{2021}+1}< \frac{2018^{2019}+1}{2019^{2019}+1}\)
\(\Rightarrow A< B\)
Ta có:
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)
\(\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}\)
\(\Rightarrow2019A=1+\frac{2019}{2019^{2021}+1}\)
\(\Rightarrow A=1+\frac{2019}{2019^{2021}+1}:2019\)
Ta lại có:
\(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
\(\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}\)
\(\Rightarrow2019B=1+\frac{2019}{2019^{2019}+1}\)
\(\Rightarrow B=1+\frac{2019}{2019^{2019}+1}:2019\)
Do \(2019^{2021}+1>2019^{2019}+1\)
\(\Rightarrow\frac{2019}{2019^{2021}+1}< \frac{2019}{2019^{2019}+1}\)
\(\Rightarrow1+\frac{2019}{2019^{2021}+1}:2019< 1+\frac{2019}{2019^{2019}+1}:2019\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
Bg
a) Ta có: A = 2011.2011 và B = 2010.2012
Xét giá trị của B:
=> B = (2011 - 1).(2011 + 1)
=> B = 2011.(2011 - 1) + 1.(2011 - 1)
=> B = 2011.2011 - 2011 + 2011 - 1
=> B = 2011.2011 - 1
Vì 2011.2011 - 1 < 2011.2011
Nên A > B
Vậy A > B.
b) Tương tự ta cũng xét giá trị của A:
=> A = (2019 - 1).(2019 + 1)
=> A = 2019.2019 - 1
Vì 2019.2019 - 1 < 2019.2019
Nên A < B
Vậy A < B
a) Ta có :
N = 2018 + 2019/2019 + 2020
= 2018/2019 + 2020 + 2019/2019 + 2020
Ta thấy : 2018/2019 + 2020 < 2018/2019 ( Vì 2019 + 2020 > 2019 )
2019/2019 + 2020 < 2019/2020 ( Vì 2019 + 2020 > 2020 )
=> 2018/2019 + 2020 + 2019/2019 + 2020 < 2018/2019 + 2019/2020
=> M > N
b) Mk ko bt làm !!
c) Ta có :
19/31 > 1/2
17/35 < 1/2
=> 19/31 > 17/35
d) Ta có :
3535/3434 = 1 + 1/3534
2323/2322 = 1 + 1/2322
Ta thấy :
1/3534 < 1/2322 ( Vì 3534 > 2322 )
=> 1 + 1/3534 < 1 + 1/2322
=> 3535/3534 < 2323/2322
Hok tốt !
Vì b > 0 => b + 2019 > 0
Ta có: \(\frac{a}{b}=\frac{a.\left(b+2019\right)}{b.\left(b+2019\right)}=\frac{a.b+a.2019}{b.\left(b+2019\right)}=\frac{a+2019}{b+2019}=\)
\(\frac{b.\left(a+2019\right)}{b.\left(b+2019\right)}=\frac{a.b+b.2019}{b.\left(b+2019\right)}\)
TH1: Nếu a < b => \(\frac{a.b+a.2019}{b.\left(b+2019\right)}< \frac{a.b+b.2019}{b.\left(b+2019\right)}\)
hay \(\frac{a}{b}< \frac{a+2019}{b+2019}\)
TH2: Nếu a = b => \(\frac{a.b+a.2019}{b.\left(b+2019\right)}=\frac{a.b+b.2019}{b.\left(b+2019\right)}\)
hay \(\frac{a}{b}=\frac{a+2019}{b+2019}\)
TH3: Nếu a > b => \(\frac{a.b+a.2019}{b.\left(b+2019\right)}>\frac{a.b+b.2019}{b.\left(b+2019\right)}\)
hay \(\frac{a}{b}=\frac{a+2019}{b+2019}\)
Xét tích : \(a(b+2019)=ab+2019a\)
\(b(a+2019)=ab+2019b\)
Vì b > 0 nên b + 2019 > 0
Nếu a > b thì \(ab+2019a>ab+2019b\)
\(a(b+2019)>b(a+2019)\)
\(\Rightarrow\frac{a}{b}>\frac{a+2019}{b+2019}\)
Nếu a < b thì \(ab+2019a< ab+2019b\)
\(a(b+2019)< b(a+2019)\)
\(\Rightarrow\frac{a}{b}< \frac{a+2019}{b+2019}\)
Nếu a = b thì rõ ràng \(\frac{a}{b}=\frac{a+2019}{b+2019}\)