dung roi

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

\(2^5=32\)

\(3^4=81\)

\(4^3=64\)

\(5^4=625\)

\(nha^{ }\)

2^5 = 32

3^4 = 81

4^3 = 64

5^4 = 625

a) 32

b) 81

c) 64

d) 625

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

a) 2³ = 8; 2⁴ = 16; 2⁵ = 32; 2⁶ = 64; 2⁷ = 128; 2⁸ = 256; 2⁹ = 512; 2¹⁰ = 1024

b) 3² = 9; 3³ = 27; 3⁴ = 81; 3⁵ = 243

c) 4² = 16; 4³ = 64; 4⁴ = 256

d) 5² = 25; 5³ = 125; 5⁴ = 625

c) 6² = 36; 6³ = 216; 6⁴ = 1296

a) 2³ = 8; 2⁴ = 16; 2⁵ = 32; 2⁶ = 64; 2⁷ = 128; 2⁸ = 256; 2⁹ = 512; 2¹⁰ = 1024

b) 3² = 9; 3³ = 27; 3⁴ = 81; 3⁵ = 243

c) 4² = 16; 4³ = 64; 4⁴ = 256

d) 5² = 25; 5³ = 125; 5⁴ = 625

e) 6² = 36; 6³ = 216; 6⁴ = 1296

HT

4 mũ 3- 2×(4- 2 mũ3)

\(25^2:100^0=\left(5^2\right)^2:1=5^4\\ 30^3:30^0=2^3.3^3.5^3:1=2^3.3^3.5^3\)

Ko biết 

`A = 2 + 2^2+ ... + 2^2017`

`=> 2A = 2^2 + 2^3 + ... + 2^2018`

`=> 2A - A = (2^2 + 2^3 + ... + 2^2018) - (2 + 2^2 + ... +2^2017)`

`=> A         = 2^2018 - 2`

`B = 1 + 3^2 + ... + 3^2018`

`=> 3^2B = 3^2 + 3^4 + ... + 3^2020`

`=> 9B-B =(3^2 + 3^4 + ... + 3^2020) - (1 + 3^2 + ... + 3^2018`

`=> 8B     = 3^2020 - 1`

`=> B       = (3^2020 - 1)/8`

`C = 5 + 5^2 - 5^3 + ... + 5^2018`

`=> 5C = 5^2 + 5^3 - 5^4 + ... +5^2019`

`=> 5C + C = ( 5^2 + 5^3 - 5^4 + ... 5^2019) + (5 + 5^2 - 5^3 + ... + 5^2018)`

`=> 6C = 55 + 5^2019`

`=> C  = (5^2019 + 55)/6`