1. x+y=xy

=> x-xy+y=0

=> x(1-y)+y=0

=> x(1-y)+y -1 =-1

=> x(1-y)- (1-y) =-1=> (1-y)(x-1)=-1

*    1-y=-1 => y=2

      x-1=1=> x=2

*     1-y =1 => y=0

       x-1=-1 => x=0

We have equation \(x+y=xy\)

\(\Rightarrow xy-x-y=0\)

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

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1=\left(-1\right).\left(-1\right)=1.1\)

So equation has two value \(\left(2;2\right),\left(0;0\right)\)

We have \(p\left(x+y\right)=xy\)

\(\Leftrightarrow xy-px-py=0\)

\(\Leftrightarrow xy-px-py+p^2=p^2\)

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

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

But p is prime so \(Ư\left(p^2\right)=\left\{1;p;p^2\right\}\)

\(\Rightarrow\left(x-p\right)\left(y-p\right)=1.p^2=p.p=p^2.1=\left(-p\right).\left(-p\right)\)

\(=\left(-1\right).\left(-p^2\right)=\left(-p^2\right).\left(-1\right)\)

So equation has values \(S=\left(p+1;p^2+p\right);\left(2p;2p\right);\left(p^2+p;p+1\right);\left(0;0\right)\)

\(;\left(p-1;p-p^2\right);\left(p-p^2;p-1\right)\)

We put \(K=\frac{2n^2+3n+3}{2n-1}\)

We have \(2n^2+3n+3=\left(2n^2-n\right)+2\left(2n-1\right)+5\)

\(=n\left(2n-1\right)+2\left(2n-1\right)+5=\left(n+2\right)\left(2n-1\right)+5\)

So \(K=n+2+\frac{5}{2n-1}\)

\(K\inℤ\Leftrightarrow5⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Prints:

So \(x\in\left\{1;0;3;-2\right\}\)then \(\frac{2n^2+3n+3}{2n-1}\in Z\)

thực sự mk rất mún giúp bn nhưng mk chưa hok tới!! xin lỗi

45646565557657767876876876565657676768876334455454655454

mình giải đc phần a) thôi:

x+y=xy
<=> x+y-xy=0
<=> x(1-y)-(1-y)+1=0
<=> (1-y)(x-1)=-1
do đó: 1-y=1;x-1=-1

 hoặc 1-y=-1; x-1=1
+) 1-y=1 => y=0

x-1=-1=> x=0

+) 1-y=-1 => y=2

x-1=1 => x=2

=> cặp x,y cần tìm là (0;0) và (2;2)

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

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.........