Ta có \(y\left(x-1\right)=x^2+2\)

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

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

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

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

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

Vì x,y nguyên nên ta có bảng

Vậy\(\left(x,y\right)=\left\{\left(4,6\right),\left(2,8\right),\left(0,2\right),\left(-2,4\right)\right\}\)thỏa mãn

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

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

\(\Leftrightarrow x\left(x-1\right)-y\left(x-1\right)+\left(x-1\right)+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=-1\\x-y+1=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=3\\x-y+1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=1\\x-y+1=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-3\\x-y+1=1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(0;-2\right),\left(4;6\right),\left(2;6\right),\left(-2;-2\right)\right\}\)

x=0 , y=0

x=2 ,  y=2

x=0 y=0

x=2 y=2

 tick nhé khi nào mk tick lại Phạm Nhật Anh

x=0   y=0

x=2   y=2

x².(y+1) + y = 30

x². (y+1) + (y+1) = 29

(y+1).(x²+1) = 29 = 1 . 29 = 29 . 1

ek cu hay qua do 

                      n.minh

Vì \(\left(2x+1\right)\left(y-3\right)=12\)

\(\Rightarrow2x+1;y-3\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

Vì \(2x+1\) là số lẻ nên \(2x+1\in\left\{-3;-1;1;3\right\}\)

Ta có bảng sau:

Vậy \(\left(x;y\right)\in\left\{\left(-2;-1\right);\left(-1;-9\right);\left(0;15\right);\left(1;7\right)\right\}\)

Ta có:

\(xy+3x-7y=21\)

\(\Rightarrow x.\left(y+3\right)-7y-21=21-21=0\)

\(x\left(y+3\right)-\left(21+7y\right)=0\)

\(x.\left(y+3\right)-7.\left(y+3\right)=0\)

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

\(\Rightarrow x-7=0\) hoặc \(y+3=0\)

TH1: x-7=0

x=0+7=7

TH2:y+3=0

y=0-3=-3

Vậy x=7; y=-3

\(xy-3x-2y=11\)

\(\Rightarrow x\left(y-3\right)-2y+6=11+6=17\)

\(\Rightarrow x\left(y-3\right)-\left(2y-6\right)=17\)

\(\Rightarrow x\left(y-3\right)-2.\left(y-3\right)=17\)

\(\Rightarrow\left(x-2\right)\left(y-3\right)=17\)

\(\Rightarrow x-2;y-3\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)

Ta có bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-15;2\right);\left(1;-14\right);\left(3;20\right);\left(19;14\right)\right\}\)

\(\Leftrightarrow4.25^x-4.5^x+1=4y^4+8y^3+12y^2+16y+41\)

\(\Leftrightarrow\left(2.5^x-1\right)^2=4y^4+8y^3+12y^2+16y+41\)

Ta có:

\(4y^4+8y^3+12y^2+16y+41=\left(2y^2+2y+2\right)^2+8y+37>\left(2y^2+2y+2\right)^2\)

\(4y^4+8y^3+12y^2+16y+41=\left(2y^2+2y+5\right)^2+4\left(y-1\right)\left(3y+4\right)\ge\left(2y^2+2y+5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}4y^4+8y^3+12y^2+16y+41=\left(2y^2+2y+3\right)^2\\4y^4+8y^3+12y^2+16y+41=\left(2y^2+2y+4\right)^2\\4y^4+8y^3+12y^2+16y+41=\left(2y^2+2y+5\right)^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y^2-y-8=0\left(\text{không có nghiệm nguyên}\right)\\8y^2-25=0\left(\text{không có nghiệm nguyên}\right)\\\left(y-1\right)\left(3y+4\right)=0\end{matrix}\right.\) 

\(\Rightarrow y=1\)

Thế vào pt ban đầu: \(25^x-5^x=20\)

Đặt \(5^x=t>0\Rightarrow t^2-t-20=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-4\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow5^x=5\Rightarrow x=1\)

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

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

Do \(xy\left(xy-1\right)\) là 2 số nguyên liên tiếp nên tích của chúng là SCP khi và chỉ khi: \(\left[{}\begin{matrix}xy=0\\xy=1\end{matrix}\right.\) 

TH1: \(xy=0\Rightarrow4x^2+9y^2=0\Rightarrow x=y=0\)

TH2: \(xy=1\Rightarrow\left(x;y\right)=\left(1;1\right);\left(-1;-1\right)\) thế vào pt đầu đều ko thỏa mãn

Lời giải:

Đặt $x+y=a; 3x+2y=b$ với $a,b\in\mathbb{Z}$ thì pt trở thành:

$ab^2=b-a-1$

$\Leftrightarrow ab^2+a+1-b=0$

$\Leftrightarrow a(b^2+1)+(1-b)=0$

$\Leftrightarrow a=\frac{b-1}{b^2+1}$

Để $a$ nguyên thì $b-1\vdots b^2+1$

$\Rightarrow b^2-b\vdots b^2+1$

$\Rightarrow (b^2+1)-(b+1)\vdots b^2+1$

$\Rightarrow b+1\vdots b^2+1$

Kết hợp với $b-1\vdots b^2+1$

$\Rightarrow (b+1)-(b-1)\vdots b^2+1$

$\Rightarrow 2\vdots b^2+1$
Vì $b^2+1\geq 1$ nên $b^2+1=1$ hoặc $b^2+1=2$
Nếu $b^2+1=1\Rightarrow b=0$. Khi đó $a=\frac{b-1}{b^2+1}=-1$
Vậy $x+y=-1; 3x+2y=0\Rightarrow x=2; y=-3$ (tm) 

Nếu $b^2+1=2\Rightarrow b=\pm 1$
Với $b=1$ thì $a=\frac{b-1}{b^2+1}=0$

Vậy $x+y=0; 3x+2y=1\Rightarrow x=1; y=-1$ (tm)

Với $b=-1$ thì $a=-1$

Vậy $x+y=-1; 3x+2y=-1\Rightarrow x=1; y=-2$ (tm)