So sánh: \(2018^{10}+2018^{11}\)và \(2019^{11}\)
K
Khách
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Những câu hỏi liên quan
2 tháng 7 2018
ta có: 201810+201811=201810.(1+2018) = 201810.2019
201911=201910.2019
=> 201810<201910 => 201810.2019 < 201910.2019
=> 201810+201811<201911
BT
2 tháng 7 2018
ta có:
201810+201811=201810.(1+2018) = 201810.2019
201911=201910.2019
=> 201810<201910 => 201810.2019 < 201910.2019
=> 201810+201811<201911
DA
18 tháng 10 2018
2019^2019 < 11x2018^2019
mk nghĩ vậy
sai thì thôi đúng thì k
mk lớp 5 thoy mà
25 tháng 1
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A
0
KT
7 tháng 1 2019
\(2018^{13}-2018^{12}=2018^{12}\left(2018-1\right)=2018^{12}.2017\)
\(2018^{11}.2018^{10}=2018^{12}.2018^9\)
Nhận thấy: \(2017< 2018^9\)=> \(2018^{12}.2017< 2018^{12}.2018^9\)
hay \(2018^{13}-2018^{12}< 2018^{11}.2018^{10}\)
CA
1
12 tháng 2 2021
Ta có: \(C=\dfrac{2019-2018}{2019+2018}\)
\(\Leftrightarrow C=\dfrac{\left(2019-2018\right)\left(2019+2018\right)}{\left(2019+2018\right)^2}\)
\(\Leftrightarrow C=\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}\)
Ta có: \(\left(2019+2018\right)^2=2019^2+2018^2+2\cdot2019\cdot2018\)
\(2019^2+2018^2=2019^2+2018^2+0\)
Do đó: \(\left(2019+2018\right)^2>2019^2+2018^2\)
\(\Leftrightarrow\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}< \dfrac{2019^2-2018^2}{2019^2+2018^2}\)
\(\Leftrightarrow C< D\)
H
1
NH
11 tháng 5 2019
\(A=\frac{2018^{2019}-1}{2018^{2019}+1}=\frac{2018^{2019}+1-2}{2018^{2019}+1}=\frac{2018^{2019}+1}{2018^{2019}+1}-\frac{2}{2018^{2019}+1}=1-\frac{2}{2018^{2019}+1}\)
\(B=\frac{2018^{2019}}{2018^{2019}+2}=\frac{2018^{2019}+2-2}{2018^{2019}+2}=\frac{2018^{2019}+2}{2018^{2019}+2}-\frac{2}{2018^{2019}+2}=1-\frac{2}{2018^{2019}+2}\)
Ta có: \(\frac{2}{2018^{2019}+1}>\frac{2}{2018^{2019}+2}\)
\(\Rightarrow1-\frac{2}{2018^{2019}+1}< 1-\frac{2}{2018^{2019}+2}\)
\(\Rightarrow A< B\)
Vậy .....
\(2018^{10}+2018^{11}=2018^{10}\cdot\left(1+2018\right)=2018^{10}\cdot2019\)
\(2019^{11}=2019^{10}\cdot2019\)
Vì \(2018^{10}< 2019^{10}\) => \(2018^{10}+2018^{11}< 2019^{11}\)
Cảm ơn bạn rất nhiều