So sánh:
\(A=\dfrac{13^{2017}+69}{13^{2019}+69}\) và \(B=\dfrac{13^{2015}+1}{13^{2017}+1}\)
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Những câu hỏi liên quan
1 so sánh
a, -36 .85.20 /25.84.34 và -30 .63.65.(-8) /117.(-200).49
b, 1985. 1987 -1/ 1980 +1985 .1986 và 1
c, A= 13^15 +1 /13^16 +1 và B= 13^16 +1/ 13^17 +1
d, 2013/2015 và 2015/2017
e, 69/68 và 72/71
So sánh A=13^2011+69/13^2013+69 và B =13^2009+1/13^2011+1
1.So Sánh
a) A=\(\dfrac{11}{2017}+\dfrac{4}{2019}và\) B=\(\dfrac{10}{2017}+\dfrac{10}{2019}\)
b) M=\(\dfrac{1}{5}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{30}+\dfrac{1}{61}+\dfrac{1}{62}và\dfrac{1}{2}\)
c) E=\(\dfrac{4116-14}{10290-35}và\) K=\(\dfrac{2929-101}{2.1919+404}\)
c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)
E = \(\dfrac{4116-14}{10290-35}\)
E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)
E = \(\dfrac{14}{35}\)
K = \(\dfrac{2929-101}{2.1919+404}\)
K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
K = \(\dfrac{29-1}{34+8}\)
K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)
Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)
\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)
\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)
Vậy E < K
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Số sánh A=13^2011+69/13^2013+69 và B=13^2009 +1/13^2011+1
Giúp tôi bài cuối cùng nhé mọi người
1)|-10|+|45|+(-|-455|)+|-755|
2)(102×132016+69×132016):132017
3)32019:(32020-24×32017)
1)
=10+45+(-455+755)
=55+300
=355.
2)
=(13^2016.169):13^2017
=13^2018:13^2017
=13.
3)
=3^2019:(3^2017.27-24.3^2017)
=3^2019:[3^2017.(27-24)]
=3^2019:3^2018
=3
A=\(\dfrac{13^{15}+1}{13^{16}+1}\) và B= \(\dfrac{13^{16}+1}{13^{17}+1}\)
so sánh A và B
\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)
B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)
vậy A=B
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\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)
ta có
\(\dfrac{13^{16}+1}{13^{17}+1}< 1\Rightarrow\dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)
vậy B<A
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\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)
ta có B<1 nên
\(\dfrac{13^{16}+1}{13^{17}+1}< \dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)
Vậy B<A
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so sánh
\(\dfrac{ }{ }\)2017/2019 và 2015/2017
\(\dfrac{2017}{2019}=1-\dfrac{2}{2019}\\ \dfrac{2015}{2017}=1-\dfrac{2}{2017}\\ Vì:\dfrac{2}{2019}< \dfrac{2}{2017}\Rightarrow1-\dfrac{2}{2019}>1-\dfrac{2}{2017}\\ \Rightarrow\dfrac{2017}{2019}>\dfrac{2015}{2017}\)
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1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
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So sánh
a, \(\dfrac{2005^{2014}+1}{2005^{2015}+1}\&\dfrac{2005^{2016}+1}{2005^{2017}+1}\)
b, \(\dfrac{19}{10}\&\dfrac{49}{40}\)
c, \(\dfrac{13}{20}\&\dfrac{33}{40}\)
ồ, lâu h ms gặp
a,
Dễ thấy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
Áp dụng khi \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)
Ta có:
\(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2016}+1+\left(2005^2-1\right)}{2005^{2017}+1+\left(2005^2-1\right)}=\dfrac{2005^{2016}+2005^2}{2005^{2017}+2005^2}=\dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)
Vậy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)
b,
\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)
Vì \(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)
c,
\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)
Vì \(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)
Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)
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Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)
\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)
\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)
\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)
\(B< A\)
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