4^32=16^16>16^15

GTNN của A=2 khi x=3

4^32=16^16

mà 16^16>16^15

suy ra 4^32>16^15

GTNN của A =2 khi x =3

Cho x/3=y/4 và x^2+y^2=100 Tìm x,y

\(32^{15}=\left(2^5\right)^{15}=2^{5.15}=2^{75}\)

\(4^{39}=\left(2^2\right)^{39}=2^{2.39}=2^{78}\)

Do \(2^{78}>2^{75}\)

\(\Rightarrow4^{39}>32^{15}\)

\(\Rightarrow1+4+4^2+...+4^{39}>32^{15}\)

\(\Rightarrow3\left(1+4+4^2+...+4^{39}\right)>32^{15}\)

Vậy \(A>B\)

mọi ng giúp mik với

A= 80.(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

A = (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

A = (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

A = (3¹⁶ - 1)(3¹⁶ + 1)(3³² + 1)

A = (3³² - 1)(3³² + 1)

A = 3⁶⁴ - 1 < 3⁶⁴ = B

=> A < B

Ta có : 

\(16^{15}=\left(4^2\right)^{15}=4^{30}\); \(4^{32}\)

Vì \(4^{30}< 4^{32}\)

=> \(16^{15}< 4^{32}\)

k mik nha

\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)

Vậy \(A< B\)

1). 4^x:16^4=32^2

=>2^2x:(2^4)^4=(2^5)^2

=>2^2x:2^16=2^10

=>2^2x=2^10.2^16

=>2^2x=2^26

=>2x=26

=>x=26:2=13

2)Ta có:

+)3^1000=(3^4)^250=81^250

+)5^750=(5^3)^250=125^250

Vì :81^250<125^250 nên 3^1000<5^750

Nó hơi dài cậu chờ tí nka !

Mình ghi nhầm đề bài 1 tí đề bài là :

So sánh 2 số A và B biết : 

A = (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1

A = (2-1)(2+1)(2^2 + 1 ) (2^4 + 1 ) ( 2^8 + 1) ( 2^16 + 1)

A = (2^2 - 1)(2^2 + 1 ) ( 2^4 + 1 )(2^8 + 1 )(2^16 + 1) 

A= ( 2^4 - 1 )( 2^4 + 1 )(2^8 + 1 )(2^16 + 1 )

A = (2^8 - 1 )(2^8 + 1 )(2^16 + 1 )

A = (2^16 - 1 )(2^16 + 1 )

A =  2^32 - 1 < 2^32 = B
Vậy A = B
k mik nka !