\(y\left(x+3\right)-5x-15=2\\ \Rightarrow y\left(x+3\right)-\left(5x+15\right)=2\\ \Rightarrow y\left(x+3\right)-5\left(x+3\right)=2\\ \Rightarrow\left(y-5\right)\left(x+3\right)=2\)

Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}y-5,x+3\in Z\\y-5,x+3\inƯ\left(2\right)\end{matrix}\right.\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(-2;7\right);\left(-1;6\right);\left(-4;3\right);\left(-5;4\right)\right\}\)

=>y.(x+3)-5(x+3)=2

=>(y-5).(x+3)=2

 Bài giải:

Ta có: x(y + 3) + 2y(x + 2) = -15

<=> xy + 3x + 2xy + 2y = -15

<=>  3xy + 3x + 2y = -15

<=> 3x(y + 1) + 2(y + 1) – 2 = -15

<=> (y + 1)(3x + 2) = -15 + 2 = -13

Ta có bảng sau:

HỌC TỐT 

#MOON#

Câu 4:

5x + 7y = 112

5(x+ y) = 112 - 2y

5(x + y) = 2(56 - y)

\(\begin{cases}x+y=2\\ 56-y=5\end{cases}\)

\(\begin{cases}x+y=2\\ y=56-5\end{cases}\)

\(\begin{cases}x=2-y\\ y=51\end{cases}\)

\(\begin{cases}x=2-51\\ y=51\end{cases}\)

\(\begin{cases}x=-49\\ y=51\end{cases}\)

Vậy (x ; y) = (-49; 51)

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

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

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

Bài 1:

xy+2x-3y=1

=>x(y+2)-3y-6=1-6

=>x(y+2)-3(y+2)=-5

=>(x-3)(y+2)=-5

=>(x-3;y+2)∈{(1;-5);(-5;1);(-1;5);(5;-1)}

=>(x;y)∈{(4;-7);(-2;-1);(2;3);(8;-3)}

Bài 1:

xy+2x-3y=1

=>x(y+2)-3y-6=1-6

=>x(y+2)-3(y+2)=-5

=>(x-3)(y+2)=-5

=>(x-3;y+2)∈{(1;-5);(-5;1);(-1;5);(5;-1)}

=>(x;y)∈{(4;-7);(-2;-1);(2;3);(8;-3)}

4:

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

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

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

1×1=3