\(2n+12⋮n-1\)

\(\Rightarrow2n-2+14⋮n-1\)

\(\Rightarrow2\left(n-1\right)+14⋮n-1\)

\(\Rightarrow14⋮n-1\)

\(\Rightarrow n-1\inƯ\left(14\right)\)

\(\Rightarrow n-1\in\left\{1;2;7;14\right\}\)

\(\Rightarrow n\in\left\{2;3;8;15\right\}\)

a. (x + 2)(3 - x) = 0

\(\Rightarrow\left\{\begin{matrix}x+2=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

b. (2x - 5)² = 9

(2x - 5)² = 3²

=> 2x - 5 = 3

2x = 3 + 5

2x = 8

x = 8 : 2

x = 4

c. (1 - 3x)³ = 8

(1 - 3x)³ = 2³

=> 1 - 3x = 2

3x = 1 - 2

3x = -1

x = \(-\frac{1}{3}\)

Mà x \(\in\) Z

=> x \(\in\phi\)

d. (x² + 1)(49.x²) = 0

\(\Rightarrow\left\{\begin{matrix}x^2+1=0\\49.x^2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x^2=1\Rightarrow x\in\phi\\x^2=0\Rightarrow x=0\end{matrix}\right.\)

Vậy x = 0

xy - 2x - 3y = 5

<=> xy - 2x - 3y + 6 = 11

<=> x(y - 2) - 3(y - 2) = 11

<=> (x - 3)(y - 2) = 11

Vậy...

xy + 3x - 7y = 21

<=> xy + 3x - 7y - 21 = 0

<=> x(y + 3) - 7(y + 3) = 0

<=> (x - 7)(y + 3) = 0

<=> \(\left[\begin{matrix}x-7=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=7\\y=-3\end{matrix}\right.\)

\(57^{1999}\)

\(=57^{1996}.57^3\)

\(=\left(57^4\right)^{499}.\overline{\left(...3\right)}\)

\(=\overline{\left(...1\right)}^{499}.\overline{\left(...3\right)}\)

\(=\overline{\left(...1\right)}.\overline{\left(...3\right)}\)

\(=\overline{...3}\)

Vậy...

a. x - (17 - 8) = 5 + (10 - 3x)

x - 9 = 5 + 10 - 3x

x - 9 = 15 - 3x

x + 3x = 15 + 9

4x = 24

x = 24 : 4

x = 6

b. 25 - (30 + x) = x - (27 - 8)

25 - 30 - x = x - 19

-5 - x = x - 19

-5 + 19 = x + x

14 = 2x

14 : 2 = x

7 = x

c. |x + 5| = 7

\(\Rightarrow\left\{\begin{matrix}x+5=7\\-\left(x+5\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\-x-5=7\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=2\\x=-12\end{matrix}\right.\)

d. x(x - 1) = 0

\(\Rightarrow\left\{\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(5n+7⋮3n+2\)

\(\Rightarrow3\left(5n+7\right)-5\left(3n+2\right)⋮3n+2\)

\(\Rightarrow15n+21-15n-10⋮3n+2\)

\(\Rightarrow11⋮3n+2\)

\(\Rightarrow3n+2\inƯ\left(11\right)\)

\(\Rightarrow3n+2\in\left\{1;-1;11;-11\right\}\)

\(\Rightarrow3n\in\left\{-1;-3;9;-13\right\}\)

\(\Rightarrow n\in\left\{-1;3\right\}\)

Vậy....

3n + 1 \(⋮\) 7

=> 3n + 1 + 14 \(⋮\) 7

=> 3n + 15 \(⋮\) 7

=> 3(n + 5) \(⋮\) 7

Vì 3 \(⋮̸\) 7 nên để 3(n + 5) \(⋮\) 7 thì n + 5 \(⋮\) 7

=> n + 5 \(\in\) B(7)

=> n + 5 = 7k (k \(\in\) N)

=> n = 7k - 5 (k \(\in\) N)

Vậy n có dạng 7k - 5 ( k \(\in\) N)