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24 tháng 6 2017

Sửa đề:

Nếu:

\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{69^{2015}+1}{69^{2017}+1}< 1\)

\(B< \dfrac{69^{2015}+1+68}{69^{2017}+1+68}\Leftrightarrow B< \dfrac{69^{2015}+69}{69^{2017}+69}\)

\(B< \dfrac{69\left(69^{2014}+1\right)}{69\left(69^{2016}+1\right)}\Leftrightarrow B< \dfrac{69^{2014}+1}{69^{2016}+1}=A\)

\(B< A\)

24 tháng 6 2017

Bạn xem đề có đúng ko đó

1 tháng 8 2017

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

Các câu còn lại tương tự

15 tháng 11 2019

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

28 tháng 4 2021

\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)

B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)

vậy A=B

28 tháng 4 2021

hơi sai sai r 

a=\(\dfrac{(13)^{15}+1}{(13)^{16}+1} =\dfrac{(13)^{15}+1}{13.(13)^{15}+1}=\dfrac{1}{3}\)

b=\(\dfrac{(13)^{16}+1}{(13)^{17}+1} =\dfrac{(13)^{16}+1}{13.(13)^{16}+1}=\dfrac{1}{13}\)

=> a=b

 

4 tháng 8 2023

\(\dfrac{2017}{2019}=1-\dfrac{2}{2019}\\ \dfrac{2015}{2017}=1-\dfrac{2}{2017}\\ Vì:\dfrac{2}{2019}< \dfrac{2}{2017}\Rightarrow1-\dfrac{2}{2019}>1-\dfrac{2}{2017}\\ \Rightarrow\dfrac{2017}{2019}>\dfrac{2015}{2017}\)

16 tháng 7 2023

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.

 

10 tháng 8 2017

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,

\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)

\(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)

c,

\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)

\(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)

Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)

10 tháng 8 2017

Áp dụng tính chất:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)

\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)

\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)

\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)

\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)

\(B< A\)