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Ta có: \(A=\frac{2020}{2021}+\frac{2021}{2022}\)
\(\Rightarrow A=\frac{2021}{2021}-\frac{1}{2021}+\frac{2022}{2022}-\frac{1}{2022}\)
\(\Rightarrow A=1-\frac{1}{2021}+1-\frac{1}{2022}\)
\(\Rightarrow A=1+1-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021\cdot2022}\)
\(B=\frac{2020+2021}{2021+2022}\)
\(\Rightarrow B=\frac{2021+2022}{2021+2022}-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{4043}\)
Vậy ta sẽ so sánh:
\(1-\frac{1}{2021\cdot2022};\frac{2}{4043}\)
Vì \(2021\cdot2022>4043\)nên \(\frac{1}{2021\cdot2022}< \frac{2}{4043}\)vậy \(1-\frac{1}{2021\cdot2022}>\frac{2}{4043}\)
\(\Rightarrow\frac{2020}{2021}+\frac{2021}{2022}>\frac{2020+2021}{2021+2022}\)
\(\Rightarrow A>B\)
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)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 \(A< B.\)
Theo phương pháp so sánh hai phân số có cùng mẫu số mà chúng ta đã
được học thì bạn Liên giải thích đúng, còn Oanh giải thích sai.
Ví dụ cho thấy bạn Oanh sai : hai phân số 3/8 và 1/2 có 3 lớn hơn 1 còn 8
lớn hơn 2 nhưng 3/8 nhỏ hơn 1/2 vì khi quy đồng về mẫu số chung là 8 thì
ta có: \(\dfrac{1}{2}=\dfrac{4}{8}>\dfrac{3}{8}\)
\(\hept{\begin{cases}3^2.\left(-2\right)^3=9.-8=-72\\-58\end{cases}}\) =>\(-72< -58=>3^2.\left(-2\right)^3< -58\)
\(\hept{\begin{cases}\left(-4\right)^3=-64\\\left|-6^2\right|=36\end{cases}=>-64< 36}=>\left(-4\right)^3< \left|-6^2\right|\)
\(3^2.\left(-2\right)^3=9.\left(-8\right)=\left(-72\right)\)
Vì (-72)<(-58) nên 32.(-2)3<(-58)
Có (-4)3 có gt âm
\(|-6^2|\)có gt dương
mà âm luôn luôn < dương
nên (-4)3<\(|-6^2|\)
mình nhầm câu b:
Áp dụng....
A=10^11-1/10^12-1<10^11-1+11/10^12-1+11=10^11+10/10^12+10=10.(10^10+1)/10.(10^11+1)
=10^10+1/10^11+1=B
Vậy A<B(câu này mới đúng còn câu b mình làm chung với câu a là sai)
a) Với a<b=>a+n/b+n >a/b
Với a>b=>a+n/b+n<a/b
Với a=b=>a+n/b+n=a/b
b) Áp dụng t/c a/b<1=>a/b<a+m/b+m(a,b,m thuộc z,b khác 0)ta có:
A=(10^11)-1/(10^12)-1=(10^11)-1+11/(10^12)-1+11=(10^11)+10/(10^12)+10=10.[(10^10)+1]/10.[(10^11)+1]
=(10^10)+1/(10^11)+1=B
Vậy A=B
ta có :\(E=\frac{2019^{2019}+1}{2019^{2020}+1}\Leftrightarrow2019\cdot E=\frac{2019^{2020}+2019}{2019^{2020}+1}=1+\frac{2019}{2019^{2020}+1}\)
\(F=\frac{2019^{2020}+1}{2019^{2021}+1}\Leftrightarrow2019\cdot F=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)
vì \(\frac{2019}{2019^{2020}+1}>\frac{2019}{2019^{2021}+1}\) nên E>F
E=2019 x 2019 x 2019 x ........ x 2019 x2019 +1 /2019 x 2019 x 2019 x.........x 2019 x 2019 + 1
E=1+1/2019+1
E=2/2020
E=1/1010
F=2019 x 2019 x 2019 x .......... x 2019 x 2019 +1 / 2019 x 2019 x 2019 x ....... x 2019 x 2019 +1
F= 1+1/2019+1
F=2/2020
F=1/1010
từ đó ta có E=F(=1/1010)
2023/2022=1+1/2022
2022/2021=1+1/2021
mà 2022>2021
nên 2023/2022<2022/2021