Bài 1: 

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

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

mà \(17^{19}+1>17^{18}+1\)

nên 17A>17B

hay A>B

2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)

\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)

mà \(98^{89}+1>98^{88}+1\)

nên C>D

Bài này có rất nhiều cách lm nhé!

Ta có : A = \(\dfrac{17^{18}+1}{17^{19}+1}\) => 17A = \(\dfrac{17^{19}+17}{17^{19}+1}\) = \(1+\dfrac{16}{17^{19}+1}\)

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

Vì \(\dfrac{16}{17^{19}+1}\) < \(\dfrac{16}{17^{18}+1}\) ( vì 17¹⁹ +1 > 17¹⁶+1 )

=> \(1+\dfrac{16}{17^{19}+1}\) < \(1+\dfrac{16}{17^{18}+1}\)

=> 17A < 17B

=> A < B ( vì 17 > 0)

Ta có :

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

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

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

\(17A=\dfrac{\left(17^{19}+1\right)+16}{17^{19}+1}\)

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

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

Lại có :

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

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

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

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

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

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

Mà : \(\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)

\(\Rightarrow1+\dfrac{16}{17^{19}+1}< 1+\dfrac{16}{17^{18}+1}\)

⇒ A < B

Vậy A < B

Giải : Vì A= \(\)\(\dfrac{17^{18}+1}{17^{19}+1}\) <1 nên áp dụng tính chất . Nếu \(\dfrac{a}{b}\) < 1 thì \(\dfrac{a}{b}\) < \(\dfrac{a+m}{b+m}\) ( a ϵ N , b ϵ N^✳ ) Ta có : A= \(\dfrac{17^{18}+1}{17^{19}+1}\)< \(\dfrac{17^{18}+1+16}{17^{19}+1+16}\) =\(\dfrac{17^{18}+17}{17^{19}+17}\) =\(\dfrac{17.17^{17}+17.1}{17.17^{18}+17.1}\)=\(\dfrac{17.\left(17^{17}+1\right)}{17.\left(17^{18}+1\right)}\) =\(\dfrac{17^{17}+1}{17^{18}+1}\) = B ⇔ A<B

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

Giải:

a) A=17¹⁸+1/17¹⁹+1

17A=17¹⁹+17/17¹⁹+1

17A=17¹⁹+1+16/17¹⁹+1

17A=1+16/17¹⁹+1

Tương tự:

B=17¹⁷+1/17¹⁸+1

17B=17¹⁸+17/17¹⁸+1

17B=17¹⁸+1+16/17¹⁸+1

17B=1+16/17¹⁸+1

Vì 16/17¹⁹+1<16/17¹⁸+1 nên 17A<17B

⇒A<B

b) A=10⁸-2/10⁸+2

    A=10⁸+2-4/10⁸+2

    A=1+-4/10⁸+2

Tương tự:

B=10⁸/10⁸+4

B=10⁸+4-4/10⁸+1

B=1+-4/10⁸+1

Vì -4/10⁸+2>-4/10⁸+1 nên A>B

c)A=20¹⁰+1/20¹⁰-1

   A=20¹⁰-1+2/20¹⁰-1

   A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-3>2/20¹⁰-1 nên B>A

⇒A<B

Chúc bạn học tốt!

Do \(\dfrac{10^{18}+1}{10^{19}+2}< 1\Rightarrow B< \dfrac{10^{18}+1+9}{10^{19}+1+9}\)

\(\Rightarrow B< \dfrac{10^{18}+10}{10^{19}+10}\)

\(\Rightarrow B< \dfrac{10\left(10^{17}+1\right)}{10\left(10^{18}+1\right)}\)

\(\Rightarrow B< \dfrac{10^{17}+1}{10^{18}+1}\)

\(\Rightarrow B< A\)

a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)

b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)

c) \(\dfrac{135}{175}=\dfrac{27}{35}\)

\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)

\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)

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

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

Vì \(\frac{16}{17^{19}+1}>\frac{16}{17^{18}+1}\Rightarrow1+\frac{16}{17^{19}+1}>1+\frac{16}{17^{18}+1}\)

\(\Rightarrow17A>17B\)

\(\Rightarrow A>B\)

Vậy A > B

A < B

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

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

Vậy \(A=B\left(=\dfrac{1}{17}\right)\).

Nếu nghĩ kĩ thì thấy bài này cũng đơn giản thôi.Thử xem cách giải của mk nè:

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

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

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

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

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

 Lại có: 17¹⁹+1>17¹⁸+1

 Suy ra:\(\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

             17A<17B

             A<B

Vậy A<B

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

\(\Rightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)

\(\Rightarrow17A=1+\frac{16}{17^{19}+1}\)

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

\(\Rightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)

\(\Rightarrow17B=1+\frac{16}{17^{18}+1}\)

\(\text{Vì }\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

\(\Rightarrow17A< 17B\)

\(\Rightarrow A< B\)

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 : 

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

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

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

\(\Leftrightarrow\frac{17^{17}+1}{17^{18}+1}\)'

\(\Rightarrow=B\)

Vậy \(A< B\)