Bài 1:

a, \(3x-5⋮2x+1\)

\(\Rightarrow2\left(3x-5\right)-3\left(2x+1\right)⋮2x+1\)

\(\Rightarrow6x-10-6x-3⋮2x+1\)

\(\Rightarrow7⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow2x\in\left\{0;-2;6;-8\right\}\)

\(\Rightarrow x\in\left\{0;-1;3;-4\right\}\)

b, \(2x-3⋮x+1\)

\(\Rightarrow2x-3-2\left(x+1\right)⋮x+1\)

\(\Rightarrow2x-3-2x-2⋮x+1\)

\(\Rightarrow1⋮x+1\)

\(\Rightarrow x+1\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Rightarrow x\in\left\{0;-2\right\}\)

c, \(3x+2⋮2x-1\)

\(\Rightarrow2\left(3x+2\right)-3\left(2x-1\right)⋮2x-1\)

\(\Rightarrow6x+4-6x+3⋮2x-1\)

\(\Rightarrow7⋮2x-1\)

\(\Rightarrow2x-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow2x\in\left\{2;0;8;-6\right\}\)

\(\Rightarrow x\in\left\{1;0;4;-3\right\}\)

d, \(2x-1⋮x+3\)

\(\Rightarrow2x-1-2\left(x+3\right)⋮x+3\)

\(\Rightarrow2x-1-2x-6⋮x+3\)

\(\Rightarrow-5⋮x+3\)

\(\Rightarrow x+3\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow x\in\left\{-2;-4;2;-8\right\}\)

Bài 2:

\(\left|x-1\right|\le2\)

\(\Rightarrow-2\le x-1\le2\)

\(\Rightarrow-2+1\le x-1+1\le2+1\)

\(\Rightarrow-1\le x\le3\)

=> x = {-1;0;1;2;3}

Bài 1:

A = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁸

3A = 3³ + 3⁴ + 3⁵ + ... + 3²⁰¹⁹

3A - A = (3³ + 3⁴ + 3⁵ + ... + 3²⁰¹⁹) - (3² + 3³ + 3⁴ + ... + 3²⁰¹⁸)

2A = 3²⁰¹⁹ - 9

A = (3²⁰¹⁹ - 9) : 2

= (3²⁰¹⁶.3³ - 9) : 2

= [ (3⁴)⁵⁰⁴.27 - 9] : 2

= [ (...1)⁵⁰⁴.27 - 9] : 2

= [ (...1).27 - 9] : 2

= [ (...7) - 9] : 2

= (....8) : 2

= ...4

Vậy c/s tận cùng của A là 4

Bài 2:

Ta có:

10¹⁹ + 10¹⁸ + 10¹⁷

= 10¹⁶.10³ + 10¹⁶.10² + 10¹⁶.10

= 10¹⁶.(10³ + 10² + 10)

= 10¹⁶.1110

= 10¹⁶.2.555

Vì 555 chia hết cho 555 nên 10¹⁶.2.555 chia hết cho 555

Vậy 10¹⁹ + 10¹⁸ + 10¹⁷ chia hết cho 555 (đpcm)

Bài 3:

x + 6 chia hết cho x + 2

=> x + 2 + 4 chia hết cho x + 2

=> 4 chia hết cho x + 2

=> x + 2 thuộc Ư(4) = {\(\pm1;\pm2;\pm4\)}

Vậy x = {-1;-3;0;-4;2;-6}

Bài 4:

Giả sử x + 4y chia hết cho 7 (1)

Vì 3x + 5y chia hết cho 7 nên 2(3x + 5y) chia hết cho 7

=> 6x + 10y chia hết cho 7 (2)

Từ (1) và (2) => (x + 4y) + (6x + 10y) chia hết cho 7

=> x + 4y + 6x + 10y chia hết cho 7

=> (x + 6x) + (4y + 10y) chia hết cho 7

=> 7x + 14y chia hết cho 7

=> 7(x + 2y) chia hết cho 7

=> Giả sử đúng

Vậy x + 4y chia hết cho 7 (đpcm)

Bài 5:

1, Ta có: \(-\left(x+2\right)^{2018}\le0\)

\(\Rightarrow-1-\left(x+2\right)^{2018}\le0\)

\(\Rightarrow A\le0\)

Dấu " = " xảy ra <=> (x + 2)²⁰¹⁸ = 0 <=> x = -2

Vậy GTNN của A là -1 khi x = -2

2, Ta có: \(x^2\ge0\)

\(\left|2y-18\right|\ge0\)

\(\Rightarrow x^2+\left|2y-18\right|\ge0\)

\(\Rightarrow-9+x^2+\left|2y-18\right|\ge-9\)

Dấu " = " xảy ra <=> \(\left\{\begin{matrix}x^2=0\\\left|2y-18\right|=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=0\\y=9\end{matrix}\right.\)

Vậy GTLN của B là -9 khi \(\left\{\begin{matrix}x=0\\y=9\end{matrix}\right.\)

Bài 6:

1, xy + 2x - y - 2 = 5

<=> x(y + 2) - (y + 2) = 5

<=> (x - 1)(y + 2) = 5

=> x - 1 và y + 2 thuộc Ư(5) = {\(\pm1;\pm5\)}

Ta có bảng:

Vậy các cặp (x;y) là (2;3) ; (0;-7) ; (6;-1) ; (-4;-3)

2, x + y = 2xy

<=> 2xy - x - y = 0

<=> 2(2xy - x - y) = 2.0

<=> 4xy - 2x - 2y = 0

<=> (4xy - 2x) - 2y - 1 = 0 - 1

<=> 2x(2y - 1) - (1 - 2y) = -1

<=> (2x - 1)(1 - 2y) = -1

=> 2x - 1 và 1 - 2y thuộc Ư(-1) = {\(\pm1\)}

Ta có bảng:

\(x^{40}=x^2\)

\(\Rightarrow x^{40}-x^2=0\)

\(\Rightarrow x^{38}.x^2-x^2=0\)

\(\Rightarrow x^2.\left(x^{38}-1\right)=0\)

\(\Rightarrow\left[\begin{matrix}x^2=0\\x^{38}-1=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0\\x^{38}=1\end{matrix}\right.\)

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

\(x^{10}=25.x^8\)

\(\Rightarrow x^{10}:x^8=25\)

\(\Rightarrow x^2=25\)

\(\Rightarrow x^2=5^2\)

\(\Rightarrow x=5\)

\(xy+2x+3y=-6\)

\(\Leftrightarrow x\left(y+2\right)+3y+6=-6+6\)

\(\Leftrightarrow x\left(y+2\right)+3\left(y+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)

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

\(A=7+7^3+7^5+...+7^{1999}\)

\(=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)

\(=\left(7+7^3\right)+7^4\left(7+7^3\right)+...+7^{1996}\left(7+7^3\right)\)

\(=350+7^4.350+...+7^{1996}.350\)

\(=350.\left(1+7^4+...+7^{1996}\right)\)

Vì \(350⋮35\) nên \(350.\left(7+7^4+...+7^{1996}\right)⋮35\)

Vậy \(A⋮35\)(đpcm)

2.

a, \(S=1+3+3^2+...+3^{49}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+2^{49}\right)\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)

\(=4+3^2.4+...+3^{48}.4\)

\(=4\left(1+3^2+...+3^{48}\right)\)

Vì \(4⋮4\) nên \(4\left(1+3^2+...+3^{48}\right)⋮4\)

Vậy \(S⋮4\) (đpcm)

c, \(S=1+3+3^2+...+3^{49}\)

\(3S=3+3^2+3^3+...+3^{50}\)

\(3S-S=\left(3+3^2+3^3+...+3^{50}\right)-\left(1+3+3^2+...+3^{49}\right)\)

\(2S=3^{50}-1\)

\(S=\left(3^{50}-1\right):2\) (đpcm)

b, Ta có: \(S=\left(3^{50}-1\right):2\)

\(=\left(3^{48}.3^2-1\right):2\)

\(=\left[\left(3^4\right)^{14}.9-1\right]:2\)

\(=\left[\overline{\left(...1\right)}^{14}.9-1\right]:2\)

\(=\left[\overline{\left(...1\right)}.9-1\right]:2\)

\(=\left[\overline{\left(...9\right)}-1\right]:2\)

\(=\overline{\left(...8\right)}:2\)

\(=\overline{...4}\)

Vậy chữ số tận cùng của S là 4

\(2n+7⋮n+1\)

\(\Rightarrow2n+2+5⋮n+1\)

\(\Rightarrow2\left(n+1\right)+5⋮n+1\)

Vì \(2\left(n+1\right)⋮n+1\) nên để \(2\left(n+1\right)+5⋮n+1\) thì \(5⋮n+1\)

\(\Rightarrow n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Vậy n = {0;-2;4;-6}

\(2n+9⋮n-3\)

\(\Rightarrow2n-6+15⋮n-3\)

\(\Rightarrow2\left(n-3\right)+15⋮n-3\)

Vì \(2\left(n-3\right)⋮n-3\) nên để \(2\left(n-3\right)+15⋮n-3\) thì \(15⋮n-3\)

\(\Rightarrow n-3\inƯ\left(15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)

Vậy n = {4;2;6;0;8;-2;18;-12}

\(\left(x^2+4x+7\right)⋮\left(x+4\right)\)

\(\Rightarrow\left[x\left(x+4\right)+7\right]⋮\left(x+4\right)\)

Vì \(x\left(x+4\right)⋮\left(x+4\right)\) nên để \(\left[x\left(x+4\right)+7\right]⋮\left(x+4\right)\) thì \(7⋮\left(x+4\right)\)

\(\Rightarrow x+4\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta có bảng

=> x = {-3;-5;3;-11}

Vậy có 4 phần tử

\(xy+y+x+1=5\)

\(\Leftrightarrow y\left(x+1\right)+\left(x+1\right)=5\)

\(\Leftrightarrow\left(y+1\right)\left(x+1\right)=5\)

=> y + 1 và x + 1 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có bảng:

Vậy các cặp (x;y) là (-2;-6) ; (0;4) ; (-6;-2) ; (4;0)

\(xy-y+x-1=7\)

\(\Leftrightarrow y\left(x-1\right)+\left(x-1\right)=7\)

\(\Leftrightarrow\left(y+1\right)\left(x-1\right)=7\)

=> y + 1 và x - 1 \(\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta có bảng:

Vậy các cặp (x;y) là (0;-8) ; (2;6) ; (-6;-2) ; (8;0)

\(xy+x+y+1=11\)

\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=11\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=11\)

=> x+1 và y+1 \(\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Ta có bảng:

Vậy các căp (x;y) là (0;10) ; (-2;-12) ; (10;0) ; (-12;-2)