Ta có 

33⁴⁴=(3.11)⁴⁴=3⁴⁴.11⁴⁴=(3⁴)¹¹.11⁴⁴=81¹¹.11⁴⁴

44³³=(4.11)³³=4³³.11³³=(4³)¹¹.11³³=64¹¹.11³³

=> 33⁴⁴>44³³

KL:

b) 5²²²²=(5²)¹¹¹¹=25¹¹¹¹

2⁵⁵⁵⁵=(2⁵)¹¹¹¹=32¹¹¹¹

=> 5²²²²<2⁵⁵⁵⁵

KL

Có 2^285 = (2^3)^95 = 8^95

     3^190 = (3^2)^95 = 9^95

Vì 8^95 < 9^95 nên 2^285 < 3^190

2²⁸⁵> 3¹⁹⁰

Ta có :

• 9999=101.99\(\Rightarrow\)9999¹⁰=(101.99)¹⁰=101¹⁰.99¹⁰
• 99²⁰=99¹⁰⁺¹⁰=99¹⁰.99¹⁰

​Vì 101¹⁰>99¹⁰\(\Leftrightarrow\)101¹⁰.99¹⁰>99¹⁰.99¹⁰\(\Leftrightarrow\)9999¹⁰>99²⁰

Vậy 9999¹⁰>99²⁰

99^20 < 9999^10

a: \(8+\dfrac{5}{13}\simeq8,\left(384615\right)< 8,415...\)

b: \(-\dfrac{4}{7}=-0.\left(571428\right)\)

tham khảo:

Ta có :2²⁷= 2^9.3=512³

            : 3¹⁸=3^6.3=729³

2²⁷ = (2³)⁹ = 8⁹

3¹⁸ = ( 3²)⁹ = 9⁹

Vì 9 > 8 nên : 9⁹ > 8⁹

Vậy suy ra: 3¹⁸ > 2²⁷

\(8^{15}=\left(2^3\right)^{15}=2^{3.15}=2^{45}\\ 16^4=\left(2^4\right)^4=2^{4.4}=2^{16}\\ 2^{45}>2^{16}\Rightarrow8^{15}>16^4\)

Cảm ơn bạn nhiều.

Chọn C

ta có : (1/8)⁵=(1/2³)⁵

               =1/2¹⁵

vì 1/2¹⁵=1/2¹⁵ => 1/2¹⁵=1/8⁵

CẢM ƠN NHIỀU NHA ^^

Ta có \(\widehat{S}+\widehat{SGQ}+\widehat{Q}=180^0\Rightarrow\widehat{S}+\widehat{Q}=180^0-\widehat{SGQ}\)

Mà \(\widehat{S}-\widehat{Q}=12^0\Rightarrow\left\{{}\begin{matrix}\widehat{S}=\dfrac{180^0-\widehat{SGQ}+12^0}{2}=96^0-\dfrac{\widehat{SGQ}}{2}\\\widehat{Q}=\dfrac{180^0-\widehat{SGQ}-12^0}{2}=84^0-\dfrac{\widehat{SGQ}}{2}\end{matrix}\right.\)

Mà GP là p/g nên \(\widehat{QGP}=\widehat{PGS}=\dfrac{\widehat{SGQ}}{2}\)

\(\Rightarrow\widehat{Q}=84^0-\widehat{QGP}\)

Ta có \(\widehat{GPS}=\widehat{Q}+\widehat{QGP}=84^0-\widehat{QGP}+\widehat{QGP}=84^0\) (tc góc ngoài)