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

\(xy-x-2y=21\)

\(\Rightarrow x\left(y-1\right)=21+2y\)

\(\Rightarrow x=\dfrac{2y+21}{y-1}\)

Vì \(x\) là số nguyên nên \(\left(2y+21\right)⋮\left(y-1\right)\)

\(\Rightarrow\left(2y-2+23\right)⋮\left(y-1\right)\)

\(\Rightarrow23⋮\left(y-1\right)\)

\(\Rightarrow y-1\inƯ\left(23\right)\)

\(\Rightarrow y-1\in\left\{1;-1;23;-23\right\}\)

\(\Rightarrow y\in\left\{2;0;24;-22\right\}\)

\(\Rightarrow x\in\left\{25;-21;3;1\right\}\)

-Vậy các cặp số \(\left(x;y\right)\) là \(\left(2;25\right)\), \(\left(0;-21\right)\), \(\left(24;-21\right)\), \(\left(-22;1\right)\).

tham khảo

=>x(2y+1)-3y-1,5=2,5

=>(y+0,5)(2x-3)=2,5

=>(2y+1)(2x-3)=5

=>\(\left(2x-3;2y+1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

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

\(2xy+x-3y=4\)

\(\Leftrightarrow4xy+2x-6y=8\)

\(\Leftrightarrow4xy+2x-6y-3=5\)

\(\Leftrightarrow2x\left(2y+1\right)-3\left(2y+1\right)=5\)

\(\Leftrightarrow\left(2x-3\right)\left(2y+1\right)=5\)

Vậy pt có các cặp nghiệm nguyên \(\left(x;y\right)=\left(-1;-1\right);\left(1;-3\right);\left(2;2\right);\left(4;0\right)\)

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

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

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

\(\Rightarrow x-1;x+1-y\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)

bảng mình xét nhầm nhé phải là như này : 

4: \(\left(x^2-1\right)\left(x^2-6\right)<0\)

TH1: \(x^2-1>0\) và \(x^2-6<0\)

=>1<\(x^2\) <6

mà x là số nguyên

nên x∈{2;-2}

TH2: \(x^2-1<0\) và \(x^2-6>0\)

=>\(x^2<1\) và \(x^2>6\)

mà 6>1

nên x∈∅

Bài 2:Sửa đề: Tìm x,y nguyên

(2x+1)(5-y)=6

=>(2x+1)(y-5)=-6

mà 2x+1 lẻ

nên (2x+1;y-5)∈{(1;-6);(-1;6);(3;-2);(-3;2)}

=>(2x;y)∈{(0;-1);(-2;11);(2;3);(-4;7)}

=>(x;y)∈{(0;-1);(-1;11);(1;3);(-2;7)}

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

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

TH1:

\(\left\{{}\begin{matrix}x-2=1\\y+1=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}x-2=-1\\y+1=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy (x;y) = (3;0); ( 1;-2)

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

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

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

Ta có bảng sau:

Vậy \(\left(x;y\right)=\left(-4;-4\right);\left(-2;0\right);\left(0;-4\right);\left(2;0\right)\)