Ta có :

\(2^{20}=2^3.2^{17}=8.2^{17}\)

\(7.2^{17}=7.2^{17}\)

Mà \(8>7\)

\(\Rightarrow8.2^{17}>7.2^{17}\Leftrightarrow2^{20}>7.2^{17}\)

2²⁰ và 7 . 2¹⁷

2²⁰ = 2³ . 2¹⁷ = 8 . 2¹⁷

Mà 8 > 7

=> 2²⁰ > 7 . 2¹⁷ 

a) 7.x - x = 5²¹ : 5¹⁹ + 3.2² - 7⁰

6x = 25 + 12 - 1

6x = 36

x = 6

b) 7x - 2x = 6¹⁷ : 6¹⁵ + 44 : 11

5x = 36 + 4

5x = 40

x = 8

c) 5x + x = 39 - 3¹¹ : 3⁹

6x = 39 - 9

6x = 30

x = 5

d) [(6x - 39) : 7]. 4 = 12

(6x - 39) : 7 = 12 : 4

(6x - 39) : 7 = 3

6x - 39 = 3 . 7

6x - 39 = 21

6x = 21 + 39

6x = 60

x = 10

x = 

(x+6)⁴=4096

(x+6)⁴=8⁴

==> x+6=8 hoặc x+6=—8

==> x=8–6 hoặc x=—8–6

==> x= 2 hoặc x=—14

2^x—3=128

2^x—3=2⁷

==> x—3=7

x=7+3

x=10

Ss: 2²⁰¹⁸ và 16⁹⁰⁰

Ta có 16⁹⁰⁰=(2⁴)⁹⁰⁰=2³⁶⁰⁰

Vì 2²⁰¹⁸<2³⁶⁰⁰

Nên 2²⁰¹⁸<2³⁶⁰⁰

17²⁰ và 4⁴⁰

Ta có 17²⁰=(17²)¹⁰=289¹⁰

4⁴⁰=(4⁴)¹⁰=256¹⁰

vì 289¹⁰>256¹⁰

Nên 17²⁰ > 4⁴⁰

Cách khác

Ta có 4⁴⁰=4^2.20=(4²)²⁰=16²⁰

Vì 16²⁰< 17²⁰

Nên 17²⁰> 4⁴⁰

a,5mũ 36=(5mũ3)mũ12=125 mũ12

11^24=(11^2)12=121^12

vì 121<125 nên 5^36>11^24

cảm ơn nha

\(2^{x-3}=128\Leftrightarrow2^{x-3}=2^7\Rightarrow x-3=7\Leftrightarrow x=10\)

\(\left(x+6\right)^4=4096\Leftrightarrow\left(x+6\right)^4=2^{12}=\left(2^3\right)^4=8^4\Rightarrow x+6=8\Leftrightarrow x=2\)

\(2^{2018}=\left(2^4\right)^{504}.2^2==16^{504}.4< 16^{900}\\ \)

\(17^{20}>16^{20}=\left(4^2\right)^{20}=4^{40}\)

\(3^{444}=\left(3^4\right)^{111}=81^{111}>64^{111}=\left(4^3\right)^{111}=4^{333}\\ \)

\(\dfrac{7^{45}}{7}=7^{44}\)

\(49.7^{20}=7^2.7^{20}=7^{22}\)

\(\Rightarrow\dfrac{7^{45}}{7}=7^{44}>7^{22}=49.7^{20}\)

7⁴⁵ : 7 ²³ =  7²² 

49 \(\times\) 7²⁰ = 7² \(\times\) 7²⁰ = 7²²

7⁴⁵ : 7²³ = 49 \(\times\) 7²⁰

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

a) ta có: 7^10 < 7^14 = (7^2)^7 = 49^7 < 50^7

=> 7^10 < 50^7

b) ta có: 5^30 = (5^3)^10 = 125^10 > 124^10

=> 5^30 > 124^10

c) ta có: 9^21 = (9^3)^7=729^7

phần d thì mk ko bk, xl bn nha

a,<

b,>

c,=

^_^