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25 tháng 4 2021

\(C=\dfrac{1957}{2007}\) và \(D=\dfrac{1935}{1985}\)

\(\Rightarrow\left\{{}\begin{matrix}C=\dfrac{1957}{2007}\\D=\dfrac{1935}{1985}\end{matrix}\right.\\\Rightarrow\left\{{}\begin{matrix}C=1-\dfrac{50}{2007}\\D=1-\dfrac{50}{1985}\end{matrix}\right. \)

Vì \(\dfrac{50}{2007}<\dfrac{50}{1985}\)

\(\Rightarrow1-\dfrac{50}{2007}>1-\dfrac{50}{1985}\\\Rightarrow C>D\)

29 tháng 4 2021

C>D

29 tháng 4 2021

C>D.

Why?

Giải:

a)Ta có:

C=1957/2007=1957+50-50/2007

                      =2007-50/2007

                      =2007/2007-50/2007

                      =1-50/2007

D=1935/1985=1935+50-50/1985

                      =1985-50/1985

                      =1985/1985-50/1985

                      =1-50/1985

Vì 50/2007<50/1985 nên -50/2007>-50/1985

⇒C>D

b)Ta có:

A=20162016+2/20162016-1

A=20162016-1+3/20162016-1

A=20162016-1/20162016-1+3/20162016-1

A=1+3/20162016-1

Tương tự: B=20162016/20162016-3

                 B=1+3/20162016-3

Vì 20162016-1>20162016-3 nên 3/20162016-1<3/20162016-3

⇒A<B

Chúc bạn học tốt!

 

 

Làm tiếp:

c)Ta có:

M=102018+1/102019+1

10M=10.(102018+1)/202019+1

10M=102019+10/102019+1

10M=102019+1+9/102019+1

10M=102019+1/102019+1 + 9/102019+1

10M=1+9/102019+1

Tương tự:

N=102019+1/102020+1

10N=1+9/102020+1

Vì 9/102019+1>9/102020+1 nên 10M>10N

⇒M>N

Chúc bạn học tốt!

9 tháng 5 2021
14 tháng 9 2023

(Sửa \(cn-bm\rightarrow cn-dm\))

Ta có :

\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)

\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)

\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)

\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)

28 tháng 7 2023

\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)

\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)

Ta có

\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)

\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)

\(\Rightarrow C>D\)

 

5 tháng 8 2023

\(a,\dfrac{-15}{17}=-1+\dfrac{2}{17}\\ -\dfrac{19}{21}=-1+\dfrac{2}{21}\\ Vì:\dfrac{2}{17}>\dfrac{2}{21}\Rightarrow-1+\dfrac{2}{17}>-1+\dfrac{2}{21}\Rightarrow-\dfrac{15}{17}>-\dfrac{19}{21}\\ b,-\dfrac{24}{35}=-1+\dfrac{11}{35};-\dfrac{19}{30}=-1+\dfrac{11}{30}\\ Vì:\dfrac{11}{35}< \dfrac{11}{30}\Rightarrow-1+\dfrac{11}{35}< -1+\dfrac{11}{30}\\ \Rightarrow-\dfrac{24}{35}< -\dfrac{19}{30}\)

a: \(\dfrac{-7}{6}=\dfrac{-7\cdot3}{6\cdot3}=\dfrac{-21}{18}\)

\(\dfrac{-11}{9}=\dfrac{-11\cdot2}{9\cdot2}=\dfrac{-22}{18}\)

mà -21>-22

nên \(-\dfrac{7}{6}>-\dfrac{11}{9}\)

b: \(\dfrac{5}{-7}=\dfrac{-5}{7}=\dfrac{-5\cdot5}{7\cdot5}=\dfrac{-25}{35}\)

\(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)

mà -25>-28

nên \(\dfrac{5}{-7}>\dfrac{-4}{5}\)

c: \(\dfrac{-8}{7}< -1\)

\(-1< -\dfrac{2}{5}\)

Do đó: \(-\dfrac{8}{7}< -\dfrac{2}{5}\)

d: \(-\dfrac{2}{5}< 0\)

\(0< \dfrac{1}{3}\)

Do đó: \(-\dfrac{2}{5}< \dfrac{1}{3}\)

NV
30 tháng 7 2021

\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)

\(\Rightarrow C>D\)