\(a)\) Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{\text{ }10}\)

Vì \(9801< 9999\)nên \(9801^{10}< 9999^{10}\)

Hay \(99^{20}< 9999^{10}\)

Vậy 99²⁰<9999¹⁰

c: \(3=\sqrt{16}-1< \sqrt{17}-1\)

a: \(11^{14}< 11^{15}\)

b: \(4^{300}=64^{100}\)

\(3^{400}=81^{100}\)

mà 64<81

nên \(4^{300}< 3^{400}\)

a, Có 3 = (3) = 9 và 2 = (2) = 8 => 3 > 2

b, Có 6 = (6) = 36 => 6 > 12

37,37 x 5959,59 = 37 x 1,01 x 59 x 101,01 = 37 x 59 x 1,01 x 101,01

59,59 x 3737,37 = 59 x 1,01 x 37 x 101,01 = 37 x 59 x 1,01 x 101,01

=> 37,37 x 5959,59 = 59,59 x 3737,37
                                              Kick mik nha

\(\dfrac{11}{-13}=-\dfrac{11}{13}=-\dfrac{13}{13}+\dfrac{2}{13}=-1+\dfrac{2}{13}\\ -\dfrac{14}{15}=-\dfrac{15}{15}+\dfrac{1}{15}=-1+\dfrac{1}{15}\)

Ta thấy : \(\dfrac{1}{15}< \dfrac{1}{13}< \dfrac{2}{13}=>-1+\dfrac{1}{15}< -1+\dfrac{2}{13}\)

hay \(\dfrac{11}{-13}>-\dfrac{14}{15}\)

a) Ta có: 3,23 < 3,32 nên -3,23 > -3,32

b) Ta có: \( - \frac{7}{3} = \frac{{ - 28}}{{12}}; - 1,25 = \frac{{ - 125}}{{100}} = \frac{{ - 5}}{4} = \frac{{ - 15}}{{12}}\)

Vì -28 < -15 nên \(\frac{{ - 28}}{{12}} < \frac{{ - 15}}{{12}}\) hay \( - \frac{7}{3}\) < -1,25

a) \(\dfrac{4}{9}< \dfrac{4}{8}=\dfrac{1}{2}\)

b) \(\dfrac{5}{8}=\dfrac{15}{24}>\dfrac{14}{24}=\dfrac{7}{12}\)

a)
4/9 < 5/10 mà 5/10 =1/2
=> 4/9<1/2
 

a: \(\dfrac{4}{9}< \dfrac{1}{2}\)

b: \(\dfrac{5}{8}>\dfrac{7}{12}\)

c: \(-\dfrac{5}{8}>-\dfrac{17}{18}\)

d: \(-\dfrac{8}{15}>-\dfrac{2}{3}\)