Question

Answer 1

`a) 100^101 = (10^2)^101 = 10^(2.101) = 10^202`

Do `202 > 201`

`=> 10^202 > 10^201`

Hay `100^101 > 10^201`

`b) (2+7)^2 = 9^2 = 81`

`2^2 + 7^2 = 4 + 49 = 53`

Mà `81 > 53`

`=>  (2+7)^2 > 2^2 + 7^2`

`c) 81^5 = (3^4)^5 = 3^(4.5) = 3^20`

Do `20 < 21`

`=> 3^20 < 3^21 `

Hay `81^5 < 3^21 `

`d) 3^40 = 3^(4.10) = (3^4)^10 = 81^10`

`5^30 = 5^(3.10) = (5^3)^10 = 125^10`

Mà `81 < 125 =>  81^10 <  125^10`

Hay `3^40 < 5^30`

Answer 2

`a)` ta có `100^101=(10^2)^101=10^202  => 10^202>10^201=> 100^101 > 10^201`

`b)` ta có` (2+7)^2=81 , 2^2+7^2=53 => 81>53 => (2+7)^2 > 2^2+7^2`

`c) `ta có  `81^5=(3^4)^5=3^20 => 81^5 < 3^21`

`d)` ta có `3^40=(3^4)^10=81^10 ,  5^30= (5^3)^10=125^10  => 3^40<5^30`