Câu 1: 3^23  >    5^12

Câu 2: 3^36 < 2^8.11^4

có 

`(-2)/3=(-2xx4)/(3xx4)=(-8)/12`

`3/(-4)=(-3)/4=(-3xx3)/(4xx3)=(-9)/12`

vì `-8> -9`

`=>(-2)/3>3/(-4)`

Ta có:
\(-\dfrac{2}{3}=-\dfrac{8}{12}\)
\(-\dfrac{3}{4}=-\dfrac{9}{12}\)
Mà \(-\dfrac{8}{12}>-\dfrac{9}{12}\)
\(\Rightarrow-\dfrac{2}{3}>-\dfrac{3}{4}\)

Ta có : \(\dfrac{2}{3}=1-\dfrac{1}{3}\)

Mặt khác, \(\dfrac{3}{4}=1-\dfrac{1}{4}\)

Mà \(\dfrac{1}{4}< \dfrac{1}{3}\) \(\Rightarrow\)\(1-\dfrac{1}{4}>1-\dfrac{1}{3}\) Hay \(\dfrac{3}{4}>\dfrac{2}{3}\) \(\Rightarrow\) \(\dfrac{-2}{3}>\dfrac{3}{-4}\)

Vậy \(\dfrac{-2}{3}>\dfrac{3}{-4}\)

a) \(\dfrac{3}{4}=0.75\)

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

\(6^2=36,2^6=64=>6^2< 2^6\)

\(2^{10}=1024>1000\)

\(6^3-4^3=216-64=152,\left(6-4\right)^3=2^3=8=>6^3-4^3>\left(6-4\right)^3\)

A)\(5^3và3^5\)

\(5^3=125;3^5=243\)

Vì\(125< 243\Rightarrow5^3< 3^5\)

B)\(4^3va3^4\)

\(4^3=64;3^4=81\)

Vì\(64< 81\Rightarrow4^3< 3^4\)

C)\(2^4va8^2\)

\(2^4=\left(2^2\right)^2=4^2\)

Vì\(4^2< 8^2\Rightarrow2^4< 8^2\)

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 \(19^4>16\cdot18\cdot20\cdot22\)

\(3^{40}=\left(3^4\right)^{10}=81^{10}\)

\(4^{30}=\left(4^3\right)^{10}=64^{10}\)

Vì \(81>64\)

`-> 81^10 > 64^10`

`-> 3^40 > 4^30`

`----`

`2^75=(2^3)^25=8^25`

`3^50=(3^2)^25=9^25`

Vì `9>8`

`-> 9^25 > 8^25`

`-> 3^50 > 2^75`.

a: 3^40=81^10

4^30=64^10

=>3^40>4^30

b: 2^75=8^25

3^50=9^25

=>2^75<3^50

a: \(2\sqrt{6}=\sqrt{24}\)

\(3\sqrt{3}=\sqrt{27}\)

mà 24<27

nên \(2\sqrt{6}< 3\sqrt{3}\)

b: \(\dfrac{2}{5}\sqrt{6}=\sqrt{\dfrac{4}{25}\cdot6}=\sqrt{\dfrac{24}{25}}\)

\(\dfrac{7}{4}\sqrt{\dfrac{1}{3}}=\sqrt{\dfrac{49}{16}\cdot\dfrac{1}{3}}=\sqrt{\dfrac{49}{48}}\)

mà 24/25<1<49/48

nên \(\dfrac{2}{5}\sqrt{6}< \dfrac{7}{4}\sqrt{\dfrac{1}{3}}\)