\(a,16^{19}=\left(2^4\right)^{19}=2^{76}\\ 8^{25}=\left(2^3\right)^{25}=2^{75}\)

Vì \(2^{76}>2^{75}=>16^{19}>8^{25}\)

b,\(3^{500}=\left(3^5\right)^{100}=243^{100}\)

Vì \(243^{100}>5^{100}=>3^{500}>5^{100}\)

cứu

a: 16^19=(2^4)^19=2^76

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

mà 76>75

nên 16^19>8^25

b: 3^500=(3^5)^100=243^100>5^100

\(a) 3^{200}=(3^2)^{100}=9^{100}\\2^{300}=(2^3)^{100}=8^{100}\)

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

\(b) 5^{40}=(5^4)^{10}=625^{10}\\3^{50}=(3^5)^{10}=243^{10}\)

Vì \(625^{10}>243^{10}\) nên \(5^{40}>3^{50}\)

#\(Toru\)

a> \(3^{200}\) và \(2^{300}\)

Ta có:\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

          \(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

Vì 9>8 nên \(9^{100}>8^{100}\)

\(\Rightarrow\)\(3^{200}>2^{300}\)

b> \(5^{40}\) và \(3^{50}\)

Ta có:\(5^{40}=5^{4.10}=\left(5^4\right)^{10}=625^{10}\)

         \(3^{50}=3^{5.10}=\left(3^5\right)^{10}=243^{10}\)

Vì 625 > 243 nên \(625^{10}>243^{10}\)

\(\Rightarrow\)\(5^{40}>3^{50}\)

`3^200=(3^2)^100=9^100`.

`2^300=(2^3)^100=8^100`.

`=> 2^300 < 3^200`.

`b, 5^40=(5^4)^10=625^10.`

`3^50=(3^5)^10=243^10`.

`=> 5^40 > 3^50`.

Ta có: `8^111 =(2^3 )^111 =2^(3.111)=2^333`

`4^170 =(2^2 )^170 =2^(2.170)=2^340`

Vì `333<340=>8^111 <4^170`

Ta có: `3^300 =3^(3.100)=(3^3 )^100=27^100`

`5^200 =5^(2.100)=(5^2 )^100 =25^100`

Vì `27>25=>3^300 >5^200`

a: 8^111=2^333

4^170=(2^2)^170=2^340

mà 333<340

nên 8^111<4^170

b: 3^300=(3^3)^100=27^100

(5^200)=(5^2)^100=25^100

mà 27>25

nên 3^300>5^200

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

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

a) 0,(26)<0,261

b) 0,15>0,14(9)

a: 0,(26)<0,261

b: 0,15>0,14(9)

a: m<n

=>2022m<2022n

b: m<n

=>-4m>-4n

a, do m<n

=> 2022m<2022n

b,do m<n

=> -4m<-4n

b)

a = 25.26 261 = 25.(26 260 +1) = 25.10.2626 + 25 = 25.10.26.101 + 25

b = 26.25 251 = 26.(25 250 + 1) = 26.10.2525 + 26 = 26.10.25.101 + 26

Suy ra a < b

a=25.26261=25.(26260+1) = 25.10.2626+25 = 25.10.26.101+25

b=26.25251=26.(25 250+1)=26.10.2525+26=26.10.25.101+26

Vì 26>25 nên b>a

a<b vì:

25<26

a) \(5^{48}=\left(5^4\right)^{12}=625^{12}\)

\(2^{108}=\left(2^9\right)^{12}=512^{12}\)

Do \(625>512\Rightarrow625^{12}>512^{12}\) \(\Rightarrow5^{48}>2^{108}\)  (1)

Lại có: \(108>105\Rightarrow2^{108}>2^{105}\)   (2)

Từ (1) và (2) \(\Rightarrow5^{48}>2^{105}\)

b) \(2^{50}=\left(2^5\right)^{10}=32^{10}\)

Do \(33>32\Rightarrow33^{10}>32^{10}\)

Vậy \(33^{10}>2^{50}\)

c) Do \(513>512\Rightarrow513^{100}>512^{100}\)   (1)

\(512^{100}=\left(2^9\right)^{100}=2^{900}\) \(=2^{10.90}=\left(2^{10}\right)^{90}=1024^{90}\) (2)

Do \(1024>1023\Rightarrow1024^{90}>1023^{90}\) (3)

Từ (1), (2) và (3) \(\Rightarrow513^{100}>1023^{90}\)

\(a,2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\\ \Leftrightarrow6+2\sqrt{2}< 3+6=9\\ b,\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}\\ 2^2=4=14-10\\ \left(2\sqrt{33}\right)^2=132>100=10^2\Leftrightarrow-2\sqrt{33}< -10\\ \Leftrightarrow\sqrt{11}-\sqrt{3}< 2\)

a: \(2\sqrt{2}< 3\)

nên \(6+2\sqrt{2}< 9\)

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

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

mà \(9^{100}>8^{100}\)

nên \(3^{200}>2^{300}\)

3^200 và 2^300

<=> (3.2)^100 và (2.3)^100

<=> 6^100 và 6^100

vậy 3^200=2^300

chúc bạn hok tốt và nhớ tick cho mk nha

\(3^{200}=\left(3^2\right)^{100}=6^{100}\)

\(2^{300}=\left(2^3\right)^{100}=6^{100}\)

\(6^{100}=6^{100}\)(\(3^{200}=2^{300}\))