Khi x=-1 thì ta sẽ có:

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

=>1+y-3-y=5

=>-2=5(vô lý)

=>x<>-1

\(x^2-xy+3x-y=5\)

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

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

=>\(y=\frac{x^2+3x-5}{x+1}\)

Để x,y là các số nguyên thì \(x^2+3x-5\) ⋮x+1

=>\(x^2+x+2x+2-7\) ⋮x+1

=>-7⋮x+1

=>x+1∈{1;-1;7;-7}

=>x∈{0;-2;6;-8}

Khi x=0 thì \(y=\frac{x^2+3x-5}{x+1}=\frac{0^2+3\cdot0-5}{0+1}=\frac{-5}{1}=-5\) (nhận)

Khi x=-2 thì \(y=\frac{x^2+3x-5}{x+1}=\frac{\left(-2\right)^2+3\cdot\left(-2\right)-5}{-2+1}=\frac{4-6-5}{-1}=\frac{-7}{-1}=7\) (nhận)

Khi x=6 thì \(y=\frac{x^2+3x-5}{x+1}=\frac{6^2+3\cdot6-5}{6+1}=\frac{36+18-5}{7}=\frac{49}{7}=7\) (nhận)

Khi x=-8 thì \(y=\frac{x^2+3x-5}{x+1}=\frac{\left(-8\right)^2+3\cdot\left(-8\right)-5}{-8+1}=\frac{64-24-5}{-7}=\frac{35}{-7}=-5\) (nhận)

x² - xy + 3x - y = 5

\(\Leftrightarrow\) x(x - y) + x - y + 2x = 5

\(\Leftrightarrow\) (x - y)(x + 1) + 2x + 2 = 7

\(\Leftrightarrow\) (x - y)(x + 1) + 2(x + 1) = 7

\(\Leftrightarrow\) (x - y + 2)(x + 1) = 7

Vì x, y \(\in\) Z nên (x - y + 2)(x + 1) \(\in\) Z

Xét các TH:

TH1: \(\left\{{}\begin{matrix}x-y+2=7\\x+1=1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2-y=7\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\y=-5\end{matrix}\right.\) (TM)

TH2: \(\left\{{}\begin{matrix}x-y+2=-7\\x+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2-y+2=-7\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\) (TM)

TH3: \(\left\{{}\begin{matrix}x-y+2=1\\x+1=7\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6-y+2=1\\x=6\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=7\end{matrix}\right.\) (TM)

TH4: \(\left\{{}\begin{matrix}x-y+2=-1\\x+1=-7\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-8-y+2=-1\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-8\\y=-5\end{matrix}\right.\) (TM)

Vậy ...

Chúc bn học tốt!

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

\(\frac{x+y}{x^2-xy+y^2}=\frac{3}{7}\)

\(\Leftrightarrow3x^2-3xy+3y^2=7x+7y\)

\(\Leftrightarrow3x^2+\left(-3y-7\right)x+3y^2-7y=0\)

Để phương trình theo nghiệm x có nghiệm thì:

\(\Delta=\left(-3y-7\right)^2-4.3.\left(3y^2-7y\right)\ge0\)

\(\Leftrightarrow0\le y\le5\)

Thế lần lược các giá trị y cái nào làm cho x nguyên thì nhận.