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

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

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

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

Bài 1 : Viết các tổng sau thành bình phương của 1 số tự nhiên 
A. 5 ³ + 6² + 8
B . 2 + 3²+ 4² + 13²

Bài 2 : So sánh các số sau 
 A . 3²⁰ và 27⁴

Ta có : 27⁴ = (3²)⁴ = 3⁸ 

Vì 20 < 8 => 3²⁰ > 27⁴

( Những câu còn lại tương tự ) - Tự làm nhé ! Mình bận ~

# Dương 

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,=

^_^

Trả lời:

a, Ta có: 3²⁰ ; 27⁴ = ( 3³ )⁴ = 3¹²

Vì 3²⁰ > 3¹² nên 3²⁰ > 27⁴

b, 2²⁵ ; 16⁶ = ( 2⁴ )⁶ = 2²⁴

Vì 2²⁵ > 2²⁴ nên 2²⁵ > 16⁶ 

Trả lời:

c, 10³⁰ = ( 10³ )¹⁰  = 1000¹⁰ ; 4⁵⁰ = ( 4⁵ )¹⁰ = 1024¹⁰

Vì 1000¹⁰ < 1024¹⁰ nên 10³⁰ < 4⁵⁰

d, 5³⁴ ; 25.5³⁰ = 5² . 5³⁰ = 5³²

Vì 5³⁴ > 5³² nên 5³⁴ > 25.5³⁰

Trả lời:

e, 10³⁰ = ( 10³ )¹⁰ = 1000¹⁰ ; 2¹⁰⁰  = ( 2¹⁰ )¹⁰ = 1024¹⁰

Vì 1000¹⁰ < 1024¹⁰ nên 10³⁰ < 2¹⁰⁰

f, 5⁴⁰ = ( 5⁴ )¹⁰ = 625¹⁰ ; 620¹⁰ 

Vì 625¹⁰ > 620¹⁰ nên 5⁴⁰ > 620¹⁰ 

Trả lời

4¹⁰⁰=2²⁰⁰

2²⁰²

Vậy 2²⁰⁰ < 2²⁰² hay 4¹⁰⁰ < 2²⁰²

3⁰ và 5⁸

3⁰ < 5⁸

a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)

\(\Rightarrow\text{ }4^{100}< 2^{202}\)

b, \(3^0=1< 5^8\)

\(3^0< 5^8\)

c, \(\left(0,6\right)^0=1\)

\(\left(-0,9\right)^6=\left(0,9\right)^6\)

\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)

d, 

e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)

\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)

Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)

a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)

\(\Rightarrow\text{ }4^{100}< 2^{202}\)

b, \(3^0=1< 5^8\)

\(3^0< 5^8\)

c, \(\left(0,6\right)^0=1\)

\(\left(-0,9\right)^6=\left(0,9\right)^6\)

\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)

d,

e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)

\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)

Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

Bài 2 :

a) \(3^2.4^3-3^2+333=3^2\left(4^3-1\right)+9.37=9.63+9.37=9\left(63+37\right)=9.100=900\)

b) \(5.4^3+24.5+41^0=20.4^2+20.6+1=20\left(16+6\right)+1=20.22+1=441\)

c) \(2^3.4^2+3^2.5-40.1^{2023}=8.16+9.5-40.1=128+45-40=128+5=133\)

a,3²⁰ và 27⁴

3²⁰=(3⁵)⁴=243⁴>27⁴

Vậy 3²⁰>27⁴

b,5³⁴ và 25x5³⁰

25x5³⁰=5²x5³⁰=5³²<5³⁴

=>5³⁴>25x5³⁰.

c,2²⁴và 26⁶

2²⁴=(2⁴)⁶=16⁶<26⁶

=>2²⁴<26⁶

d,10³⁰và 4⁵⁰

10³⁰=(10³)¹⁰=1000¹⁰

4⁵⁰=(4⁵)¹⁰=1024¹⁰

Vì 1000¹⁰<1024¹⁰nên 10³⁰<4⁵⁰.

e,2³⁰⁰và 3²⁰⁰

2³⁰⁰=(2³)¹⁰⁰=8¹⁰⁰

3²⁰⁰=(3²)¹⁰⁰=9¹⁰⁰

Vì 8¹⁰⁰<9¹⁰⁰ nên 2³⁰⁰<3²⁰⁰

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)