=>x(y+1)-2y-2+5=0

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

=>\(\left(x-2;y+1\right)\in\left\{\left(1;-5\right);\left(-5;1\right);\left(-1;5\right);\left(5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;-6\right);\left(-3;0\right);\left(1;4\right);\left(7;-2\right)\right\}\)

(2x + y) + (2y - x)i = (x - 2y + 3) + (y + 2x + 1)i

\(y\left(x+1\right)^2=-x^2+2018x-1\)

\(\Leftrightarrow y=\dfrac{-x^2+2018x-1}{\left(x+1\right)^2}=-1+\dfrac{2020x}{\left(x+1\right)^2}\)

\(\Rightarrow\dfrac{2020x}{\left(x+1\right)^2}\in Z\)

Mà x và \(x\left(x+2x\right)+1\) nguyên tố cùng nhau

\(\Rightarrow2020⋮\left(x+1\right)^2\)

Ta có 2020 chia hết cho đúng 2 số chính phương là 1 và 4

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=1\\\left(x+1\right)^2=4\end{matrix}\right.\) \(\Rightarrow x=\left\{0;1\right\}\) \(\Rightarrow y\)

b.

Từ pt đầu:

\(x^2+xy-2y^2+2\left(x-y\right)=0\)

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

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

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

Thế xuống dưới ...

a: (2x+1)(y-3)=10

mà 2x+1 lẻ

nên (2x+1;y-3)∈{(1;10); (5;2)}

=>(2x;y)∈{(0;13); (4;5)}

=>(x;y)∈{(0;13); (2;5)}

b: (3x-2)(2y-3)=1

=>(3x-2;2y-3)∈{(1;1);(-1;-1)}

=>(3x;2y)∈{(3;4);(1;2)}

=>(x;y)∈{(1;2);(1/3;1)}

mà x là số tự nhiên

nên (x;y)∈{(1;2)}

c: (x+1)(2y-1)=12

mà 2y-1 lẻ

nên (x+1;2y-1)∈{(12;1);(4;3)}

=>(x;2y)∈{(11;2);(3;4)}

=>(x;y)∈{(11;1);(3;2)}

d: x+6=y(x-1)

=>y(x-1)=x-1+7

=>y(x-1)-(x-1)=7

=>(x-1)(y-1)=7

=>(x-1;y-1)∈{(1;7);(7;1)}

=>(x;y)∈{(2;8);(8;2)}

e: x-3=y(x+2)

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

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

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

=>(x+2;y-1)∈{(5;-1)}(Vì x+2>=2 khi x là số tự nhiên)

=>(x;y)∈{(3;0)}

BÀi 2:

Đặt x = 11...1(n chữ số 1), khi đó

a = x

b = 100..05(n-1 chữ số 0) = 100...00(n chữ số 0) + 5

b = 99...9(n chữ số 9) + 1 + 5 = 9x +6

=> \(ab+1=x\left(9x+6\right)+1\)

=> \(ab+1=9x^2+6x+1=\left(3x+1\right)^2\)

Vậy ab + 1 là 1 số chính phương

Ta có: \(x^4-2x^2y^2+y^4\)

\(=\left(x^2-y^2\right)^2\)

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

\(-2x^3-2y^3\)

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

\(=-2\left\lbrack\left(x+y\right)^3-3xy\left(x+y\right)\right\rbrack=-2\left\lbrack1^3-3xy\cdot1\right\rbrack\)

=-2+6xy

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

=1-2xy

Ta có: \(A=x^4-2x^3-2x^2y^2-2y^3+x^2+y^2+y^4\)

\(=\left(x^4-2x^2y^2+y^4\right)+\left(-2x^3-2y^3\right)+\left(x^2+y^2\right)\)

\(=x^2+y^2-2xy-2+6xy+1-2xy=x^2+2xy+y^2-1\)

\(=\left(x+y\right)^2-1=1^2-1=0\)