\(31^{15}< 32^{15}=\left(2^5\right)^{15}=2^{75}\)

\(17^{19}>16^{19}=\left(2^4\right)^{19}=2^{76}\)

Do \(2^{75}< 2^{76}\Rightarrow31^{15}< 2^{75}< 2^{76}< 17^{19}\)

Vậy \(31^{15}< 17^{19}\)

a)=

b)>

c)=

Đầu đề hơi sai 

a,Ta có: \(1-\frac{17}{19}=\frac{2}{19}\) và \(1-\frac{15}{17}=\frac{2}{17}\)

Vì \(\frac{2}{19}< \frac{2}{17}\) nên \(\frac{17}{19}< \frac{15}{17}\)

a. 17/19 = 19/19 -2/19 = 1- 2/19

15/17 = 17/17 - 2/17 = 1 - 2/17

Vì 2/17 >2/19 nên 1-2/17 < 1 - 1/19

hay 15/17 < 17/19

17/15>29/32

12/18<13/17

16/51>31/96

21/25>19/29>60/81

\(6-\sqrt{17}=\sqrt{36}-\sqrt{17}\)

Với : 

\(\sqrt{36}-\sqrt{17}>\sqrt{31}-\sqrt{17}\)

Mặt khác : 

\(\sqrt{31}-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

Nên : 

\(6-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

Cách khác:

Ta có: \(\left(\sqrt{31}-\sqrt{19}\right)^2=50-2\sqrt{589}\)

\(\left(6-\sqrt{17}\right)^2=53-12\sqrt{17}=50+3-12\sqrt{17}\)

mà \(-2\sqrt{589}< 3-12\sqrt{17}\)

nên \(\sqrt{31}-\sqrt{19}>6-\sqrt{17}\)

\(6-\sqrt{17}=\sqrt{36}-\sqrt{17}>\sqrt{31}-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

HAY  \(6-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

đợi xíu mik hỏi mẹ

a) >

b) >

c) <

d) <

\(17^{20}=17^{\left(4.5\right)}=\left(17^4\right)^5=83521^5\)

\(31^{15}=31^{\left(3.5\right)}=\left(31^3\right)^5=29791^5\)

Vì \(83521^5< 29791^5\Rightarrow17^{20}>31^{15}\)

\(17^{20}>31^{15}\)

chuc bn hoc gioi!

nhae

~~~~~~~~~~~~~~~

chuc bn hoc gioi!

!!!

Ta có :

\(17^{20}=17^{\left(5.4\right)}=\left(17^4\right)^5=83521^5\)

\(31^{15}=31^{\left(5.3\right)}=\left(31^3\right)^5=29791^5\)

Mà : 83521>29791

=> 83521⁵>29791⁵