`#3107.101107`

`a)`

Ta có:

\(\dfrac{2727}{3131}=\dfrac{2727\div27}{3131\div31}=\dfrac{27}{31}\)

Vì \(\dfrac{27}{31}=\dfrac{27}{31}\)

\(\Rightarrow\dfrac{27}{31}=\dfrac{2727}{3131}\)

`b)`

Ta có:

\(\dfrac{11}{31}=1-\dfrac{20}{31}=1-\dfrac{200}{310}\)

\(\dfrac{111}{311}=1-\dfrac{200}{311}\)

Vì \(\dfrac{200}{310}>\dfrac{200}{311}\)

\(\Rightarrow1-\dfrac{200}{310}< 1-\dfrac{200}{311}\)

\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}.\)

27/31 = 2727/3131

11/31 bé hơn 111/311

\(a,\dfrac{2727}{3131}=\dfrac{2727:101}{3131:101}=\dfrac{27}{31}\\ Vậy:\dfrac{27}{31}=\dfrac{2727}{3131}\)

\(\dfrac{2727}{3131}=\dfrac{27.101}{31.101}=\dfrac{27}{31}\)

⇒\(\dfrac{27}{31}=\dfrac{2727}{3131}\)

\(\dfrac{27}{31}=\dfrac{27\times101}{31\times101}=\dfrac{2727}{3131}\)

\(\dfrac{11}{31}\) và \(\dfrac{111}{311}\)

\(\dfrac{11}{31}\) = \(\dfrac{11\times10}{31\times10}\) = \(\dfrac{110}{310}\) = 1 - \(\dfrac{200}{310}\) 

\(\dfrac{111}{311}\) = 1 - \(\dfrac{200}{311}\)  

Vì \(\dfrac{200}{310}\) > \(\dfrac{200}{311}\)

Nên \(\dfrac{11}{31}\)  < \(\dfrac{111}{311}\)

\(\dfrac{11}{31}=\dfrac{11x311}{31x311}=\dfrac{3421}{31x311}\)

\(\dfrac{111}{311}=\dfrac{111x31}{31x311}=\dfrac{3441}{31x311}\)

mà \(\dfrac{3421}{31x311}< \dfrac{3441}{31x311}\)

\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}\)

31¹¹ < 32¹¹ = (2⁵)¹¹ = 2⁵⁵

17¹⁴ > 16¹⁴ = (2⁴)¹⁴ = 2⁵⁶

Vì 31¹¹ < 2⁵⁵ < 2⁵⁶ < 17¹⁴ 

=> 31¹¹ < 17¹⁴

ko bt làm thì có :))

a)17¹⁴>16^14₌2⁵⁶>32¹¹=2²²>31¹¹

b)1023³⁰<1024³⁰=2³⁰⁰<2³⁰⁵=32⁶¹<33⁶¹

c)82¹⁷>81¹⁷=3⁶⁸>3⁶³=27²¹>26²¹

d)3³⁹<3⁴²=9²¹<11²¹

\(A=\frac{19^{30}+5}{19^{31}+5}=>19A=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\left(1\right)\)

\(B=\frac{19^{31}+5}{19^{32}+5}=>19B=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\left(2\right)\)

từ (1) and (2)

=>19A>19B

=>A>B