\(=2^3.\left(19-14\right)+1=2^3.5+1=8.5+1=40+1=41\)

cảm ơn bn 

A) Ta có: 

\(\dfrac{12}{13}=\dfrac{13}{13}-\dfrac{1}{13}=1-\dfrac{1}{13}\)

\(\dfrac{13}{14}=\dfrac{14}{14}-\dfrac{1}{14}=1-\dfrac{1}{14}\)

Mà \(1-\dfrac{1}{13}< -\dfrac{1}{14}\)

\(\Rightarrow\dfrac{12}{13}< \dfrac{13}{14}\)

B) Ta có:

\(\dfrac{125}{251}=\dfrac{251}{251}-\dfrac{126}{251}=1-\dfrac{126}{251}\)

\(\dfrac{127}{253}=\dfrac{253}{253}-\dfrac{126}{253}=1-\dfrac{126}{253}\)

Mà: \(1-\dfrac{126}{251}< 1-\dfrac{126}{253}\)

\(\Rightarrow\dfrac{125}{251}< \dfrac{127}{253}\)

\(5^6:5^4+2^3-1^{2023}=5^{6-4}+8-1=5^2+8-1=25+8-1=33-1=32\)

32

a, 18.(-24) +  35.42 - 18.76  + 35.(-142)

=  [ 18.(-24) - 18.76] + (35.42 + 35(-142)]

= -18.[ 24 + 76]  - 35.( 142 - 42)

= -18.100  - 35. 100

= - 100.(18 + 35)

= - 100. 53

= - 5300

h,   104 : (-13) - [56 - 220 : (-4)]

  = 104 : (-13)  - [ 56 + 55]

= -8 - 111

= - 119 

\(3^2\cdot5\left[7:1^{2022}-\left(11-9\right)^2\right]+2022^0\)

\(=9\cdot5\cdot\left(7:1-2^2\right)+1\)

\(=9\cdot5\cdot\left(7-4\right)+1\)

\(=45\cdot3+1\)

\(=135+1\)

\(=136\)

a) \(\dfrac{7}{12}< \dfrac{7+1}{12+1}< \dfrac{78}{13}\Rightarrow\dfrac{7}{12}< \dfrac{8}{13}\)

b) \(-4,25=-\dfrac{425}{100}=-\dfrac{17}{4}=-\dfrac{34}{8}< -\dfrac{28}{8}\Rightarrow-4,25< -\dfrac{28}{8}\)

c) \(-0,33>-0,5=-\dfrac{1}{2}=-\dfrac{19}{38}\Rightarrow-0,33>-\dfrac{19}{38}\)

d) \(\dfrac{11}{13}< \dfrac{11+2}{13+2}=\dfrac{13}{15}\Rightarrow\dfrac{11}{13}< \dfrac{13}{15}\Rightarrow-\dfrac{11}{13}>-\dfrac{13}{15}\)

Ai phúp tui làm -49 phần 211 và 13 phần 1999 

Và bài 311 phần 256 và 199 và 203 

26 phần 27 và 96 phần 97

\(A=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)

\(B=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)

mà \(\sqrt{12}+\sqrt{11}< \sqrt{14}+\sqrt{13}\)

nên A>B