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

cho mình bài c với đc ko?mình ko bik làm😫😖

tham khảo:

https://olm.vn/hoi-dap/detail/56325533229.html

3.13579/34567 = 40737/34567 = 34567+6170/34567

3.13580/34569 = 40740/34569 = 34569+6171/34569

vì : 34567+6170/34567 < 34569+6171/34569

nên: 3.13579/34567 < 3.13580/34569

vậy: 13579/34567 < 13580/34569

B=\(\dfrac{10^9+1}{10^{10}+1}< \dfrac{10^5+1+9}{10^{10}+1+9}=\dfrac{10^9+10}{10^{10}+10}=\dfrac{10.\left(10^8+1\right)}{10\left(10^9+1\right)}\)

= A

Ta có F < 1 nên

\(\dfrac{13580}{34569}< 1< \dfrac{13580+\left(-1\right)}{34569+\left(-1\right)}=\dfrac{13579}{34568}\)<\(\dfrac{13579}{34567}\)

Từ đó suy ra \(\dfrac{13580}{34569}< \dfrac{13579}{34567}\)hay\(\dfrac{13579}{34567}>\dfrac{13580}{34569}\)

Vậy E > F

2/

a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)

\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)

-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)

b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)

-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)

c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)

\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)

-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)

a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)

c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)

Câu b thì gg search nhé

Ta có: \(A=\dfrac{3^{10}+1}{3^9+1}\)

\(\Leftrightarrow A=\dfrac{3^{10}+3-2}{3^9+1}\)

hay \(A=3-\dfrac{2}{3^9+1}\)

Ta có: \(B=\dfrac{3^9+1}{3^8+1}\)

\(\Leftrightarrow B=\dfrac{3^9+3-2}{3^8+1}\)

hay \(B=3-\dfrac{2}{3^8+1}\)

Ta có: \(3^9+1>3^8+1\)

\(\Leftrightarrow\dfrac{2}{3^9+1}< \dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}>-\dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}+3>-\dfrac{2}{3^8+1}+3\)

hay A>B

a. 19/10 > 10/11

b. 11/10 = 12/11

c. 9/10 = 10/11

a)\(\dfrac{19}{10}>\dfrac{10}{11}\)

b)\(\dfrac{11}{10}=\dfrac{12}{11}\)

c)\(\dfrac{9}{10}< \dfrac{10}{11}\)

a) \(< \)

b) \(>\)

c) \(< \)

d) \(>\)

e) \(< \)

g) \(>\)

h) \(>\)

k) \(>\)