Bạn tham khảo nhé 

Ta có công thức : 

\(\frac{a}{b}>\frac{a+c}{b+c}\) \(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(C=\frac{100^{90}+1}{100^{80}+1}>\frac{100^{90}+1+99}{100^{80}+1+99}=\frac{100^{90}+100}{100^{80}+100}=\frac{100\left(100^{89}+1\right)}{100\left(100^{79}+1\right)}=\frac{100^{89}+1}{100^{79}+1}=D\)

Vậy \(C>D\)

Chúc bạn học tốt ~

THANK YOU cho mik gửi lời mời kết bạn nhé !

a) A < B              

b) C > D

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!

1) Phân tích A ra :

 A= 17¹⁷.17+\(\frac{1}{17^{18}.17}\)+1 So sánh với B ta có: A có 17¹⁸>17¹⁷ của B nhưng B lại có 1/17¹⁸>1/17¹⁹.

Mà 17¹⁸>1/17¹⁸ nên suy ra A>B

2) Bài nay tương tự bài trên. 

2/(2012+2013) < 2/(2012 + 2012) = 2/ (2.2012) = 1/2012 
2009/(2012+2013) < 2009/2012 

=> 2011/(2012+2013) = 2/(2012+2013) + 2009/(2012+2013) < 1/2012 + 2009/2012 
=> 2011/(2012+2013) < 2010/2012 (a) 

2012/(2012+2013) < 2012/2013 (b) 

lấy (a) + (b) => (2011+2012)/(2012+2013) < 2010/2012 + 2012/2013 

vậy B < A 

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

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

Ta có

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

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

\(\Rightarrow C>D\)

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

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

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

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

\(\Rightarrow C>D\)

\(C=5^{2018}+\frac{1}{5^{2017}+1}=\left(5^{2017}+1\right)+\frac{1}{5^{2017}+1}\)

\(D=5^{2018}+\frac{1}{5^{2018}+1}=\left(5^{2017}+1\right)+\left(1+\frac{1}{5^{2017}+2}\right)\)

Do \(\frac{1}{5^{2017}+1}< 1+\frac{1}{5^{2017}+2}\)

Nên \(C< D\)

Ta có : C = \(\frac{5^{2018}+1}{5^{2017}+1}\)

=> \(\frac{C}{5}=\frac{5^{2018}+1}{5^{2018}+5}=1-\frac{4}{5^{2018}+5}\)

Lại có D = \(\frac{5^{2019}+1}{5^{2018}+1}\)

=> \(\frac{D}{5}=\frac{5^{2019}+1}{5^{2019}+5}=1-\frac{4}{5^{2019}+5}\)

Vì \(\frac{4}{5^{2018}+5}>\frac{4}{5^{2019}+5}\Rightarrow1-\frac{4}{5^{2018}+5}< 1-\frac{4}{5^{2019}+5}\Rightarrow\frac{C}{5}< \frac{D}{5}\Rightarrow C< D\)