Đáp án:
a, Ta có:
`63^15 < 64 ^ 15 = (2^6) ^15 = 2 ^90`
`34^18 > 32 ^18 =(2^5)^18 = 2^90`
`⇔ 34^18> 2 ^90 > 63^15`
Hay `34^ 18> 63^15`
b, Ta có :
`83^9 > 81^9 = (3^4)^9 = 3^36`
`26^12 < 27^12 = (3^3)^12 = 3^36`
`⇔ 83^9 > 3^36 > 26^12`
Hay `83^9> 26^12`
`text{ @toanisthebest}`