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

a: \(\left(x^2+2\right)\left(x+3\right)>0\)

mà \(x^2+2\ge2>0\forall x\)

nên x+3>0

=>x>-3

mà x nguyên

nên x∈{-2;-1;0;1;2;...}

b: (x+3)(x-4)<0

Th1: \(\begin{cases}x+3>0\\ x-4<0\end{cases}\Rightarrow\begin{cases}x>-3\\ x<4\end{cases}\)

=>-3<x<4

mà x nguyên

nên x∈{-2;-1;0;1;2;3}

TH2: \(\begin{cases}x+3<0\\ x-4>0\end{cases}\Rightarrow\begin{cases}x<-3\\ x>4\end{cases}\)

=>Loại

c: \(3x^2+12x>0\)

=>3x(x+4)>0

=>x(x+4)>0

=>\(\left[\begin{array}{l}x>0\\ x<-4\end{array}\right.\)

mà x nguyên

nên x∈Z\{-4;-3;-2;-1;0}

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 \(\Leftrightarrow x=y=\frac{1}{2}\)