Câu 1: 3^23  >    5^12

Câu 2: 3^36 < 2^8.11^4

1) \(5^{199}< 5^{200}=25^{100}\)

\(3^{300}=27^{100}>25^{100}\)

\(\Rightarrow3^{300}>5^{199}\)

\(\Rightarrow\dfrac{1}{3^{300}}< \dfrac{1}{5^{199}}\)

2)  a) \(107^{50}=\left(107^2\right)^{25}=11449^{25}\)

\(73^{75}=\left(73^3\right)^{25}=389017^{25}>11449^{25}\)

\(\Rightarrow107^{50}< 73^{75}\)

b) \(54^4< 5^{12}< 21^{12}\Rightarrow54^4< 21^{12}\)

Giúp mình với

chắc là >

a) \(\dfrac{3}{4}=0.75\)

b) \(-2\dfrac{1}{2}=\dfrac{-5}{2}< \dfrac{-5}{3}\)

\(1+\dfrac{1}{3}+1\dfrac{3}{4}\div2\dfrac{1}{2}+\dfrac{2}{3}\)

\(=1+\dfrac{1}{3}+\dfrac{7}{4}\div\dfrac{5}{2}+\dfrac{2}{3}\)

\(=1+\dfrac{1}{3}+\dfrac{7}{10}+\dfrac{2}{3}\)

\(=\left(1+\dfrac{7}{10}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{17}{10}+1\)

\(=\dfrac{27}{10}\)

Bài 1:

a) \(\dfrac{-17}{36}\) và \(\dfrac{23}{-48}\) 

\(\dfrac{-17}{36}=\dfrac{-17.4}{36.4}=\dfrac{-68}{144}\) 

\(\dfrac{23}{-48}=\dfrac{-23}{48}=\dfrac{-23.3}{144.3}=\dfrac{-69}{144}\) 

Vì \(\dfrac{-68}{144}>\dfrac{-69}{144}\) nên \(\dfrac{-17}{36}>\dfrac{23}{-48}\) 

b) \(\dfrac{-1}{3}\) và \(\dfrac{2}{5}\) 

Vì \(\dfrac{-1}{3}\) là số âm mà \(\dfrac{2}{5}\) là số dương nên \(\dfrac{-1}{3}< \dfrac{2}{5}\) 

c) \(\dfrac{2}{7}\) và \(\dfrac{5}{4}\) 

Vì \(\dfrac{2}{7}< 1\) mà \(\dfrac{5}{4}>1\) nên \(\dfrac{2}{7}< \dfrac{5}{4}\) 

d) \(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\) 

\(\dfrac{267}{-268}=\dfrac{-267}{268}=\dfrac{-267.449}{268.449}=\dfrac{-119883}{120332}\) 

\(\dfrac{-1347}{1343}=\dfrac{-1347.89}{1343.89}=\dfrac{-119883}{119527}\) 

Vì \(\dfrac{-119883}{120332}>\dfrac{-119883}{119527}\) nên \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

Bài 2:

\(\dfrac{5}{2}-\left(1\dfrac{3}{7}-0,4\right)=\dfrac{5}{2}-\dfrac{10}{7}-\dfrac{2}{5}=\dfrac{47}{70}\) 

\(\left(1+2+3+4\right)^2=\left(1+2+3+4\right)\left(1+2+3+4\right)=1^2+1\cdot2+1\cdot3+1\cdot4+2\cdot1+2^2+2\cdot3+2\cdot4+3\cdot1+3\cdot2+3^2+3\cdot4+4\cdot1+4\cdot2+4\cdot3+4^2=1^2+2^2+3^2+4^2+1\cdot2+1\cdot3+1\cdot4+2\cdot1+2\cdot3+2\cdot4+3\cdot1+3\cdot2+3\cdot4+4\cdot1+4\cdot2+4\cdot3>1^2+2^2+3^2+4^2\)

bằng nhau

h mình nha

bạn chép sai đề sao lai 1/3+1/4+1/5+..........1/2dc

a: \(\left(1+2+3+4\right)^2=10^2=100\)

\(1^3+2^3+3^3+4^3=1+8+27+64=100\)

Do đó: \(\left(1+2+3+4\right)^2=1^3+2^3+3^3+4^3\)

b: \(19^4=130321\)

\(16\cdot18\cdot20\cdot22=126720\)

mà 130321>126720

nên \(19^4>16\cdot18\cdot20\cdot22\)