Theo đề ra, ta có:

425 : x dư 29\(\Rightarrow396⋮x\)

857 : x dư 32\(\Rightarrow825⋮x\)

\(\Rightarrow x\inƯC\left\{396;825\right\}\)

Ta có:

\(396=2^2.3^2.11\)

\(825=3.5^2.11\)

\(\RightarrowƯCLN\left(396;825\right)=3.11=33\)

\(\RightarrowƯC\left(396;825\right)=\left\{\pm1;\pm3;\pm11;\pm33\right\}\)

Mà \(x\inℕ\Rightarrow x\in\left\{1;3;11;33\right\}\)

Vì 1003:(1002-x) > 0

=> 2014 -1003:(1002-x) > 2014 -0

=> A > 2014

=> giá trị NN của A = 2015

Đáp án : ........................................................................................................................................................................................................................................................................................................................................................................................................................................................

:)))

Tích 2 số: a.b = BCNN(a,b).ƯCLN(a,b) = 40500
Vì ƯCLN(a,b) = 15 => a = 15k; b = 15q (với (k,q) = 1)
=> 15k.15q = 40500 => k.q = 180
Vì (k, q) = 1 => (k, q) ∈ {(4,45); (5,36), (9,20); (20;9); (36;5);(45;4)}
Vậy (a, b) ∈ {(60;675);(75;540);(135;300);(300;135);(540;75);(675;60)}

Theo đề ra, ta có:

\(\hept{\begin{cases}\left(a+7\right)⋮28\\\left(a+7\right)⋮24\\\left(a+7\right)⋮16\end{cases}}\Rightarrow\left(a+7\right)\in BC\left(28;24;16\right)\)

Ta có:

\(28=2^2.7\)

\(24=2^3.3\)

\(16=2^4\)

\(\Rightarrow BCNN\left(16;18;24\right)=2^4.3.7=336\)

\(\Rightarrow\left(a+7\right)=BC\left(16;18;24\right)=\left\{0;336;672;1008;...\right\}\)

Mà đề ra a là số nhỏ nhất có bốn chữ số

\(a+7=1008\Rightarrow a=1008-7\Rightarrow a=1001\)

A = 1111111111 – 22222=1111x10001-2x1111=1111(10001-2)=1111x9999=1111.3.3333=3333x3333=3333^2

=>..............

Ta có

\(1111111111=\frac{9999999999}{9}=\frac{9999999999+1-1}{9}=\frac{10^{10}-1}{9}.\)

\(22222=2.11111=\frac{2.99999}{9}=\frac{2\left(99999+1-1\right)}{9}=\frac{2\left(10^5-1\right)}{9}\)

\(\Rightarrow A=\frac{10^{10}}{9}-\frac{1}{9}-2.\frac{10^5}{9}+\frac{2}{9}=\)

\(=\left(\frac{10^5}{3}\right)^2-2.\frac{10^5}{3}.\frac{1}{3}+\left(\frac{1}{3}\right)^2=\left(\frac{10^5}{3}-\frac{1}{3}\right)^2\)

a) Ta có : 2^300=2^3.100=8^100

               3^200=3^2.100=9^100

Ta thấy 8^100<9^100 

=>2^300<3^200

b)Ta có:54^4=(2.3^3)^4=2^4.3^12

            21^12=(3.7)^12=3^12.7^17

Ta thấy 3^12=3^12

            2^4<7^12

Do đó 3^12.2^4<3^12.7^13

Hay 54^4<21^12

c) Ta có 5^100=5^100

             2^200=(2^2)^100=4^100

Ta thấy 5^100>4^100

Do đó 5^100>2^200

d)Ta có 10^20=(10^2)^10=20^10

Ta thấy 20^10<40^10

Hay 10^20<40^10

a) 5.125.625 = 5.5³.5⁴ = 5^{1 + 3 + 4} = 5⁸

b) 10.100.1000 = 10.10².10³ = 10^{1 + 2 + 3} = 10⁶ 

 c) 8⁴.16⁵.32 = (2³)⁴.(2⁴)⁵.2⁵ = 2^3.4.2^4.5.2⁵ = 2^{3.4 + 4.5 + 5} = 2^{12 + 20 + 5} = 2³⁷

d) 27⁴.81¹⁰ = (3³)⁴.(3⁴)¹⁰ = 3^3.4.3^4.10 = 3^{3.4 + 4.10} = 3^{12 + 40} = 3⁵²

Viết các tích sau dưới dạng lũy thừa:

\(a)\) \(5.125.625=5.5^3.5^4=5^8\)

\(b)\)\(10.100.1000=10.10^2.10^3=10^6\)

\(c)\)\(8^4.16^5.32=\left(2^3\right)^4.\left(2^4\right)^5.2^5=2^{12}.2^{20}.2^5=2^{37}\)

\(d)\)\(27^4.81^{10}=\left(3^3\right)^4.\left(3^4\right)^{10}=3^{12}.3^{40}=3^{52}\)

A = 137 . 454 + 206

    = 137 . (453 + 1) + 206

    = 137 . 453 + 137 + 206

    = 137 . 453 + 343

    = 137 . 453 + (453 - 110)

    = 137 . 453 + 453 - 110

    = 138 . 453 - 110

=>A=B

sai bép học giốt

\(A=137.454+206\)\(=137\left(453+1\right)+206\)\(=137.453+137.206\)\(=137.453+343\)

\(B=453.138-110\)\(=453\left(137+1\right)-110\)\(=453.137+453-110\)\(=453.137-343\)

Ta có: \(137.453.343=453.137+343\)

Vậy A = B

Trả lời 

a)

Cách 1  :   \(A=\left\{1;2;3;4;5;....;99\right\}\)

Cách 2 :   \(A=\left\{x\in N,x< 100\right\}\)