\(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

duyệt đi

\(\left(\frac{1}{32}\right)^2=\frac{1^7}{32^7}=\frac{1}{32^2}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{35}}\)

\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{36}}\)

Vì: 2³⁵ < 2³⁶ nên \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)

\(\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{35}}\)

\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{36}}\)

Vì \(2^{35}\frac{1}{2^{36}}\)

Vậy \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

(1/32)^7 = (1/2^5)^7 =(1/2)^35 > ( 1/2 ) ^36 = (1/2^4 )^9 = ( 1/ 16 ) ^9

(1\32)^7 =(1\2)^35 >(1\2)^36=(1\16)^9 
=>(1\32)^7 > (1\16)^9 

Vậy (1\32)^7 > (1\16)^9 

7 = 3 + 4 = √9 + √16

Do 10 > 9 nên √10 > √9

17 > 16 nên √17 > √16

⇒ √10 + √17 > √9 + √16

Vậy √10 + √17 > 7

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(1/8)²³ = 1/(2³)²³ = 1/2⁶⁹

(1/32)¹⁶ = 1/(2⁵)¹⁶ = 1/2⁸⁰

Do 69 < 80 nên 2⁶⁹ < 2⁸⁰

⇒ 1/2⁶⁹ > 1/2⁸⁰

Vậy (1/8)²³ > (1/³²)¹⁶

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5 = √25

Do 27 > 25 nên √27 > √25

Vậy √27 > 5

Vậy A = B

tích mik nha mik se cho lời giải

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

A = 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1 

 =(2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)+1

=(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)+1

=(2⁸-1)(2⁸+1)(2¹⁶+1)+1

=(2¹⁶-1)(2¹⁶+1)+1

=2³²-1+1

=2³² = B

vậy A=B