so sánh qua số trunhg gian

\(32^{81}< 31^{90}\)

\(32^{81}>31^{90}\)

32⁸¹ - 31⁹⁰ = -x

=> 32⁸¹ < 31⁹⁰ 

~ Hok tốt ~

333⁴⁴⁴ và 444³³³

ta có : 333⁴⁴⁴ = ( 333⁴ )¹¹¹ =12296370321¹¹¹

444³³³ = ( 444³ )¹¹¹ =  87528384¹¹¹

^vì 12296370321 >  87528384

=> 333⁴⁴⁴ > 444³³³

Bài 1:

333^444 = (333^4)^111

444^333 = (444^3)^111

333^4 = 3^4.111^4 = 81.111^4

444^3 = 4^3.111^3 = 64.111^3 < 81.111^4

444^3< 333^4

(333^4)^111 > (444^3)^111

333^444 > 444^333

Ta có: \(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19.\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

    \(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19.\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

Nên  \(19A< 19B\Rightarrow A< B\)

Nhầm: Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow A>B\)

a) 31¹¹ < 32¹¹=(2⁵)¹¹=2⁵⁵

    17¹⁴>16¹⁴=(2⁴)¹⁴=2⁵⁶

=> 17¹⁴>2⁵⁶>2⁵⁵>31¹¹

b) 125³¹=(5³)³¹=5⁹³

    25⁴⁷=(5²)⁴⁷=5⁹⁴>5⁹³=125³¹

c) A=1+2²+2³+...+2¹⁰⁰=2¹⁰¹-1 < 2¹⁰¹=B

d) C=1990¹⁰+1990⁹=1990⁹(1990+1)=1991.1990⁹ < 1991.1991⁹=1991¹⁰

=> C < 1991¹⁰=B