vì 2017¹⁰⁰ + 1 < 2017¹⁰¹ + 1

\(\Rightarrow\frac{2017^{100}+1}{2017^{101}+1}< \frac{2017^{100}+1+2016}{2017^{101}+1+2016}=\frac{2017^{100}+2017}{2017^{101}+2017}=\frac{2017.\left(2017^{99+1}\right)}{2017.\left(2017^{100}+1\right)}=\frac{2017^{99}+1}{2017^{100}+1}\)

Vậy \(\frac{2017^{99}+1}{2017^{100}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)

so sánh 2 phân số cùng mẫu thì ta xét tử

đừng nói không làm được chứ

có cùng mẫu đâu bạn ơi

\(A=\frac{2017^{99}}{2017^{100}-2}\) 

=> \(2017A=\frac{2017^{100}}{2017^{100}-2}=\frac{2017^{100}-2+2}{2017^{100}-2}=1+\frac{2}{2017^{100}-2}\)​

\(B=\frac{2017^{100}}{2017^{101}-2}\)

=>\(2017B=\frac{2017^{101}}{2017^{101}-2}=\frac{2017^{101}-2+2}{2017^{101}-2}=1+\frac{2}{2017^{101}-2}\)​

Do \(\frac{2}{2017^{100}-2}>\frac{2}{2017^{101}-2}\)

Nên 2017A > 2017B

Vậy A > B

 ta có :

\(25^{1008}=\left(5^2\right)^{1008}=5^{2.1008}=5^{2016}\)

mà \(5^{2017}>5^{2016}\)

\(\Rightarrow\)\(5^{2017}>\left(5^2\right)^{1008}\)

\(\Rightarrow\)\(5^{2017}>25^{1008}\)

có \(5^{2017}=\left(5^2\right)^{1008}\times5\)\(=25^{1008}\times5\)

mà \(=25^{1008}\times5\)> \(25^{1008}\)

nên \(5^{2017}>25^{1008}\)

Ta có:

\(5^{2017}>5^{2016}=\text{[}5^2\text{]}^{1008}=25^{1008}\)

Suy ra: 5²⁰¹⁷ > 25¹⁰⁰⁸

Ta có:

\(1-A=1-\frac{10^{101}-1}{10^{102}-1}=\frac{10^{102}-1-\text{[}10^{101}-1\text{]}}{10^{102}-1}=\frac{10^{102}-1-10^{101}+1}{10^{102}-1}\)\(=\frac{10^{102}-10^{101}}{10^{102}-1}=\frac{10^{101}\left[10-1\right]}{10^{101}\text{[}10-\frac{1}{10^{101}}\text{]}}=\frac{10-1}{10-\frac{1}{10^{101}}}=\frac{9}{10-\frac{1}{10^{101}}}\)

\(1-B=1-\frac{10^{100}+1}{10^{101}+1}=\frac{10^{101}+1-\left[10^{100}+1\right]}{10^{101}+1}=\frac{10^{101}+1-10^{100}-1}{10^{100}+1}\)

\(=\frac{10^{101}-10^{100}}{10^{101}+1}=\frac{10^{100}\left[10-1\right]}{10^{100}\text{[}10+\frac{1}{10^{100}}\text{]}}=\frac{10-1}{10+\frac{1}{10^{100}}}=\frac{9}{10+\frac{1}{10^{100}}}\)

Vì \(\frac{9}{10-\frac{1}{10^{101}}}>\frac{9}{10+\frac{1}{10^{100}}}\Rightarrow A< B\)

Trả lời

Ko chép lại đề

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Trả lời :

a)\(\frac{99}{100}< 1\)và \(\frac{100}{99}>1\)nên \(\frac{99}{100}< \frac{100}{99}\)

~ Hok tốt ~

b,....đề....

ta có :1-\(\frac{99}{100}\)=\(\frac{1}{100}\) 

         1-\(\frac{100}{101}\)= \(\frac{1}{101}\) 

mà \(\frac{1}{100}\) >    \(\frac{1}{101}\) 

 => \(\frac{99}{100}\) >   \(\frac{100}{101}\)

2016^100+2016^99>2017^100

cái phép tính 1 lớn hơn

2016^100+2016^99= 2016^99.2016^1+2016^99.1=2016^99.(2016+1)=2016^99.2017

2017^100= 2017^99.2017

Do 2016^99.2017< 2017^99.2017

Nên 2016^100+2016^99< 2017^100