Tham khảo :

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

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

Vì \(17^{19}>17^{18}=>17^{19}+1>17^{18}+1\)

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

\(=>17A< 17B=>A< B\)

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

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!

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

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

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

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

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

a, \(\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{1}{20}\)

b, \(-\dfrac{24}{17}+\dfrac{7}{17}+\dfrac{1}{16}=\dfrac{-17}{17}+\dfrac{1}{16}=-1+\dfrac{1}{16}=-\dfrac{15}{16}\)

\(\dfrac{14}{13}+\left(\dfrac{-1}{13}-\dfrac{19}{20}\right)=\left(\dfrac{14}{13}+\dfrac{-1}{13}\right)-\dfrac{19}{20}=\\ \dfrac{13}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}=\dfrac{1}{20}\)

Câu b em làm tương tự nhé

\(a.\dfrac{14}{13}+\left(\dfrac{-1}{13}+\dfrac{19}{20}\right)=\left(\dfrac{14}{13}+\dfrac{-1}{13}\right)-\dfrac{19}{20}=1+\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}=\dfrac{1}{20}\\ b.\dfrac{-24}{17}+\left(\dfrac{-7}{17}-\dfrac{1}{16}\right)=\left(\dfrac{-24}{17}+\dfrac{7}{17}\right)+\dfrac{1}{16}=-1+\dfrac{1}{16}=\dfrac{-16}{16}+\dfrac{1}{16}=\dfrac{-15}{16}\)

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

\(\dfrac{16}{17}:1\dfrac{1}{17}:1\dfrac{1}{18}:1\dfrac{1}{19}\)

\(=\dfrac{16}{17}:\dfrac{18}{17}:\dfrac{19}{18}:\dfrac{20}{19}=\dfrac{16}{17}.\dfrac{17}{18}.\dfrac{18}{19}.\dfrac{19}{20}=\dfrac{16}{20}=\dfrac{4}{5}\)

Ta có :

\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)

\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)