2^59=2^17.2^42

10^17=2^17.5^17

5^17<5^18=25^4<32^4<32^8=2^40<2^42

=> 5^17<2^42

vậy ....

a) 2018 + (– 3)  <  2018

          b) (– 105) + 5 > (– 105)

          c) (– 59) + (– 10) < (–59)

Ta có 2⁵⁹ < 2⁶⁰ = (2⁴)¹⁵ = 16¹⁵ < 17¹⁵

Vậy 2⁵⁹ < 17¹⁵

17^15 > 2^59 .                                             

Có : 17^15 > 16^15 = (2^4)^15 = 2^60

Vì 2^60 > 2^59 mà 17^15 > 2^60 => 17^15 > 2^59

k mk nha bạn

2⁵⁹ > 10¹⁷

2⁵⁹ < 10¹⁷

\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)

Nhận thầy 10⁸ - 1 > 10⁸ - 3

=> \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)

=> \(1+\frac{3}{10^8-1}< \frac{3}{10^8-3}+1\)

=> A < B

b) 17C = \(\frac{17\left(17^{203}+1\right)}{17^{204}+1}=\frac{17^{204}+1+16}{17^{204}+1}=1+\frac{16}{17^{204}+1}\)

17D = \(\frac{17\left(17^{202}+1\right)}{17^{203}+1}=\frac{17^{203}+1+16}{17^{203}+1}=1+\frac{16}{17^{203}+1}\)

Nhận thầy 17²⁰³ + 1 < 17²⁰⁴ + 1

=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)

=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\)