a) \(2x^2-3xy-2y^2=2\)

\(\Rightarrow2x^2+xy-4xy-2y^2=2\)

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

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

\(\Rightarrow\left(2x+y\right);\left(x-2y\right)\in\left\{-1;1;-2;2\right\}\)

Ta giải các hệ phương trình sau với x;y nguyên 

1) \(\left\{{}\begin{matrix}2x+y=-1\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-2\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}2x+y=1\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}4x+2y=2\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}5x=4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}2x+y=-2\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-4\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

4)  \(\left\{{}\begin{matrix}2x+y=2\\x-2y=1\end{matrix}\right.\) \(\left\{{}\begin{matrix}4x+2y=4\\x-2y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(1;1\right)\right\}\)

b) \(xy-y+x=9\)

\(\Rightarrow y\left(x-1\right)+x-1+1=9\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=8\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;-9\right);\left(2;7\right);\left(-1;-5\right);\left(3;3\right);\left(-3;-3\right);\left(5;1\right);\left(-7;-2\right);\left(9;0\right)\right\}\)

Đoạn:

2x
2 + 2y
2 − 3z
2= -100 là như thế nào bạn nhỉ?

Bạn viết lại đề để mọi người hiểu hơn nhé.

\(a,xy-x-y=2\\ x\left(y-1\right)-y=2\\ x\left(y-1\right)-y+1=2+1\\ x\left(y-1\right)-\left(y-1\right)=3\\ \left(y-1\right)\left(x-1\right)=3\\ Th1:x-1=-1=>x=0\\ y-1=-3=>y=-2\\ Th2:x-1=-3 =>x=-2\\ y-1=-1=> y=0\\ Th3:x-1=3=> x=4\\ y-1=1=>y=2\\ Th4:x-1=1=>x=2\\ y-1=3=>y=4\)

Vậy......

\(b,2x^2+3xy-2y^2=7\\ 2x^2+\left(4xy-xy\right)-2y^2=7\\ x\left(2x-y\right)+2y\left(2x-y\right)=7\\ \left(2x-y\right)\cdot\left(x+2y\right)=7\)

Nếu 2x-y=1; x+2y = 7

=> 2(2x-y) + x + 2y = 9

=> 4x - 2y + x +2y = 9

=> (4x+x) + (2y-2y) = 9

=> 5x + 0 = 9 

=> x = 9/5 (ktm)

Nếu 2x-y=7; x+2y = 1

=> 2(2x-y) + x+ 2y = 15

=> 4x - 2y + x +2y =15

=> (4x +x)+ (2y-2y) =15

=> 5x +0 =15

=> x= 3 (tm)

=> y= -1 (Tm)

Nếu  2x-y=-7; x+2y = -1

=> 2(2x-y) + x+ 2y = -15

=> 4x - 2y + x +2y =-15

=> (4x +x)+ (2y-2y) =-15

=> 5x +0 =-15

=> x= -3 (tm)

=> y= 1 (tm)

Nếu 2x-y=-1 ; x+2y = -7

=> 2(2x-y) + x+ 2y = -9

=> 4x - 2y + x +2y = -9

=> (4x +x)+ (2y-2y) =-9

=> 5x +0 =-9

=> x= -9/5 (ktm)

=> y= -1

Vậy.........

\(x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3\left(x+y\right)+10=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10=5^3-2.5^2+3.5+10=100\)

Answer:

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

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

Mà x và y nguyên \(\Rightarrow x^2+5⋮3x-2\)

\(\Rightarrow9\left(x^2+5\right)⋮3x-2\)

\(\Rightarrow9x^2-6x+6x-4+49⋮3x-2\)

\(\Rightarrow49⋮3x-2\)

\(\Rightarrow3x-2\in\left\{\pm49;\pm7;\pm1\right\}\)

\(\Rightarrow3x\in\left\{51;9;3;-5;1;-47\right\}\)

\(\Rightarrow x\in\left\{1;3;7\right\}\)

Trường hợp 1: Với \(x=1\) ta thay vào (1)

\(\Rightarrow y=6\)

Trường hợp 2: Với \(x=3\) ta thay vào (1)

\(\Rightarrow y=2\)

Trường hợp 3: Với \(x=7\)ta thay vào (1)

\(\Rightarrow y=6\)

ai giúp em bài1 và phần b bài 2 với ạ