a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

mà 17^19+1>17^18+1

nên A<B

b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)

\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)

2^2021-1<2^2022-1

=>1/2^2021-1>1/2^2022-1

=>-1/2^2021-1<-1/2^2022-1

=>C<D

A=(17^18+1)/(17^19+1)

17A=17(17^18+1)/17^19+1=17^19+17/17^19+1

17A=(17^19+1)+16/(17^19+1)=1+16/17^19+1    

B=(17^17+1)/(17^18+1)

17B=17(17^17+1)/17^18+1=17^18+17/17^18+1

17B=(17^18+1)+16/(17^18+1)=1+16/17^18+1

Từ (1) và (2)⇒1+16/17^19+1<1+16/17^18+1

=> 17A<17B

Hay A<B

Vậy A<B

Bài 1:

Ta thấy A < 1

=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)

Vậy A < B

Bài 2:

Ta thấy C < 1

=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)

Vậy C < D

Ta có:

\(A=\frac{17^{18}+1}{17^{19}+1}\)

\(17A=\frac{17\left(17^{18}+1\right)}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}\)

\(17A=\frac{(17^{19}+1)+16}{(17^{19}+1)}=1+\frac{16}{17^{19}+1}\)          (1)

\(B=\frac{17^{17}+1}{17^{18}+1}\)

\(17B=\frac{17\left(17^{17}+1\right)}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}\)

\(17B=\frac{(17^{18}+1)+16}{(17^{18}+1)}=1+\frac{16}{17^{18}+1}\)          (2)

Từ (1) và (2) => \(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

=>\(17A< 17B\)

Hay \(A< B\)

Vậy \(A< B\)

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 : 

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

 Ta co
A=17.17^17+1/17.17^18+1
   =1+(17^17+1/17^18+1)
Vi B=17^17+1/17^18+1
=>.B<A
 
chuan lun

Ta có : \(17^{17}-2< 17^{18}-2\)

Mà mẫu số càng lớn thì p/s càng bé

\(\Rightarrow\)\(\frac{2}{17^{17}-2}< \frac{2}{17^{18}-2}\)

Lại có :\(17^{18}< 17^{19}\)

\(\Rightarrow\)\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)​​\(17^{18}-\frac{2}{17^{17}-2}< 17^{19}-\frac{2}{17^{18}-2}\)( Vì số bị trừ càng lớn thì hiệu càng bé )

ta có A=\(\frac{17^{18}+1}{17^{19}+1}\)<\(\frac{17^{18}+1+16}{17^{19}+1+16}\) (nếu a/b<1 thì a+c/b+c>a/b)

A<\(\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}\)

A,<\(\frac{17^{17}+1}{17^{18}+1}\)=B

hay A<B

\(A=\frac{17^{18}+1}{17^{19}+1}\) với \(B=\frac{17^{17}+1}{17^{18}+1}\)

Ta có :B=\(\frac{17^{17}+1}{17^{18}+1}=\frac{17^{18}+17}{17^{19}+17}\)

Ta có:1-B=\(1-\frac{17^{18}+17}{17^{19}+17}=\frac{17^{19}+17-17^{18}-17}{17^{19}+17}=\frac{17^{19}-17^{18}}{17^{19}+17}\)

         1-A=1-\(\frac{17^{18}+1}{17^{19}+1}=\frac{17^{19}+1-17^{18}-1}{17^{19}+1}=\frac{17^{19}-17^{18}}{17^{19}+1}\)

Do \(17^{19}+1< 17^{19}+10\Rightarrow1-A>1-B\)

                  \(\Rightarrow A< B\)