a: \(x^2+y^2(x-y+1)-\left(x-1\right)\cdot y=22\)

=>\(x^2 + y^2(x - y + 1) - xy + y = 22\)

=>\(x^2-xy+y^2(x-y+1)+y=22\)

=>\(x(x-y+1)-x+y^2(x-y+1)+y=22\)

\(\iff(x-y+1)(x+y^2)-(x-y)=22\)

\(\iff(x-y+1)(x+y^2)-(x-y+1)+1=22\)

=>\((x-y+1)(x+y^2-1)=21\)

TH1: \(\begin{cases}x-y+1=1\\ x+y^2-1=21\end{cases}\)

=>\(\begin{cases}x=y\\ y+y^2-1=21\end{cases}\iff y^2+y-22=0\)

=>\(y^2+y+\frac14-\frac{89}{4}=0\)

=>\(\left(y+\frac12\right)^2=\frac{89}{4}\)

=>\(\left[\begin{array}{l}y+\frac12=\frac{\sqrt{89}}{2}\\ y+\frac12=-\frac{\sqrt{89}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}y=\frac{\sqrt{89}-1}{2}\left(loại\right)\\ y=\frac{-\sqrt{89}-1}{2}\left(loại\right)\end{array}\right.\)

=>Loại

TH2: \(\begin{cases}x-y+1=21\\ x+y^2-1=1\end{cases}\iff\begin{cases}x=y+20\\ (y+20)+y^2-1=1\end{cases}\iff y^2+y+18=0\)

=>\(y^2+y+\frac14+\frac{71}{4}=0\)

=>\(\left(y+\frac12\right)^2+\frac{71}{4}=0\) (vô lý)

=>Loại

TH3: \(\begin{cases}x-y+1=3\\ x+y^2-1=7\end{cases}\iff\begin{cases}x=y+2\\ (y+2)+y^2-1=7\end{cases}\iff y^2+y-6=0\)

=>(y+3)(y-2)=0

=>y=-3 hoặc y=2

Nếu y=-3 thì x-y+1=3

=>x-y=2

=>x-(-3)=2

=>x+3=2

=>x=-1

Nếu y=2 thì x-y+1=3

=>x-2+1=3

=>x-1=3

=>x=4

TH4: \(\begin{cases}x-y+1=7\\ x+y^2-1=3\end{cases}\iff\begin{cases}x=y+6\\ (y+6)+y^2-1=3\end{cases}\iff y^2+y+2=0\)

=>y∈∅

=>Loại

 Nếu \(x< -3\) thì \(x^2+x+3< x^2\) và \(x^2+x+3>\left(x+1\right)^2\), vô lý.

 Nếu \(x>2\) thì \(x^2+x+3>x^2\) và \(x^2+x+3< \left(x+1\right)^2\), cũng vô lý.

 Do đó \(x\in\left\{-3;-2;-1;0;1;2\right\}\)

 Thử từng giá trị, ta thấy \(\left(x;y\right)\in\left\{\left(-3;3\right);\left(-3;-3\right)\right\}\) là các cặp số thỏa ycbt.

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

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

Ta thấy :

\(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\ge\left|x+2+1-x\right|=3\)

\(\left(y+2\right)^2\ge0\Rightarrow3-\left(y+2\right)^2\le3\)

\(\Rightarrow VT\ge3\ge VP\)

Để \(VP=VT\Leftrightarrow\hept{\begin{cases}\left|x+2\right|+\left|x-1\right|=3\\3-\left(y+2\right)^2=3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-2;-1;0;1\\y=-2\end{cases}}\)

Vậy các cặp (x;y) nguyên là (-2;-2) ; (-1;-2) ; (0;2) ; (1;2)

Hello

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

=>\(\hept{\begin{cases}x-2=1\\x+y-2=3\end{cases};\hept{\begin{cases}x-2=-1\\x+y-2=-3\end{cases};\hept{\begin{cases}x-2=3\\x+y-2=1\end{cases};\hept{\begin{cases}x-2=-3\\x+y-2=-1\end{cases}}}}}\)

=> \(\hept{\begin{cases}x=3\\y=2\end{cases};\hept{\begin{cases}x=1\\y=-2\end{cases};\hept{\begin{cases}x=5\\y=-2\end{cases};\hept{\begin{cases}x=-1\\y=2\end{cases}}}}}\)

Ta có \(y^2=3-2\left|2x+3\right|\ge0\Leftrightarrow0\le\left|2x+3\right|\le\dfrac{3}{2}\)

Mà \(x,y\in Z\Leftrightarrow\left|2x+3\right|\in\left\{0;1\right\}\)

Với \(\left|2x+3\right|=0\Leftrightarrow x=-\dfrac{3}{2}\left(loại\right)\)

Với \(\left|2x+3\right|=1\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\Leftrightarrow y^2=1\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)

Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-1;1\right);\left(-1;-1\right);\left(-2;1\right);\left(-2;-1\right)\)

ta có  \(\left(x-2\right)^3=x^3-6x^2+12x-8>x^3-6x^2+12x-27=y^3\)

ta có \(6x^2-12x+27>0vớimoix\)

\(=>-6x^2+12x-27< 0\)

\(=>y^3>x^3\)

mà x y nguyên nên y^3 nguyên =>\(y^3=\left(x-1\right)^3\)

|X-3|+|y-5|=0

x^2y là sao bạn hình như sai ở chỗ đó

đó là (x^2)*y nha