\(a,2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Vì \(8^{100}< 9^{100}\) nên \(2^{300}< 3^{200}\)

\(b,8^5=32768\)

\(6^6=46656\)

Vì \(32768< 46656\) nên \(8^5< 6^6\)

\(c,3^{450}=\left(3^3\right)^{150}=27^{150}\)

\(5^{300}=\left(5^2\right)^{150}=25^{150}\)

Vì \(27^{150}>25^{150}\) nên \(3^{450}>5^{300}\)

#Ayumu

a) Ta có :\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Leftrightarrow2\sqrt{5}< 5\)

b) Ta có : \(\dfrac{16}{9}< 12\Rightarrow\sqrt{\dfrac{16}{9}}< \sqrt{12}\Leftrightarrow\dfrac{1}{3}\cdot\sqrt{16}< \sqrt{12}\)

a: \(2\sqrt{5}=\sqrt{20}\)

\(5=\sqrt{25}\)

mà 20<25

nên \(2\sqrt{5}< 5\)

b: \(\dfrac{1}{3}\cdot\sqrt{16}=\sqrt{\dfrac{1}{9}\cdot16}=\sqrt{\dfrac{16}{9}}\)

\(\sqrt{12}=\sqrt{\dfrac{108}{9}}\)

mà 16<9

nên \(\dfrac{1}{3}\sqrt{16}< \sqrt{12}\)

a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!

a)

\(\dfrac{-2}{3}\)>\(\dfrac{5}{-8}\)

b)

\(\dfrac{398}{-412}\)<\(\dfrac{-25}{-137}\)

c)

\(\dfrac{-14}{21}\)<\(\dfrac{60}{72}\)

a <

b <

c <

a)

5/−8

b)

−25/−137

c)

60/72

b: \(\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{7}+2-\sqrt{7}+\sqrt{3}=2+\sqrt{3}\)

a: 99^20=9801^10<9999^10

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

`a)`

  `m > n`

`<=>2m > 2n`

`<=>2m+3 > 2n+3`

Vậy `2n+3 < 2m+3`

_________________________

`b)`

   `m > n`

`<=>-m < -n`

`<=>-m-5 < -n-5`

Vậy `-n-5 > -m-5`

a)\(m>n\Rightarrow2m>2n\Rightarrow2m+3>2n+2\)

b)\(m>n\Rightarrow-m< -n\Rightarrow-m-5< -n-5\)

a)2^6=8^2

b)5^3<3^5

a) \(2^6\) và \(8^2\)

   \(2^6=\left(2^2\right)^3\)

  \(8^2=\left(2^3\right)^2\)\(=2^6\)

 \(\Rightarrow\) \(2^6=8^2\)