Điều kiện \(x\ne\pm3;y\ne-2\):

 \(P=\frac{2x+3y}{xy+2x-3y-6}-\frac{6-xy}{xy+2x+3y+6}-\frac{x^2+9}{x^2-9}.\)

=> \(P=\frac{2x+3y}{\left(y+2\right)\left(x-3\right)}-\frac{6-xy}{\left(y+2\right)\left(x+3\right)}-\frac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{\left(2x+3y\right)\left(x+3\right)-\left(6-xy\right)\left(x-3\right)-\left(x^2+9\right)\left(y+2\right)}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{2x^2+3xy+6x+9y-6x+x^2y+18-3xy-x^2y-9y-2x^2-18}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{0}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}=0\)

=> P=0 (với mọi x khác 3, -3 và y khác -2)

Ta có : xy - 2x + 3y - 5 = 0 

<=> x(y - 2) + 3y - 6 + 1 = 0

<=> x(y - 2) + 3(y - 2) + 1 = 0

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

Suy ra :  (y - 2) (x + 3) thuộc Ư(-1) = {-1;1}

Th1 : nếu y - 2 = -1 thì x + 3 = -1 => y = 1 ; x = -4

Th2 : nếu y - 2 = 1 thì x + 3 = 1 => y = 3 , x = -2 

what the hell???

avatar mèo đen

g: (x+3y)(x-3y+2)

=(x+3y)(x-3y)+2(x+3y)

=x^2-9y^2+2x+6y

h: (x+2y)(x-2y+3)

=(x+2y)(x-2y)+3(x+2y)

=x^2-4y^2+3x+6y

i: (x^2-xy+y^2)(x+y)

=x^3+x^2y-x^2y-xy^2+xy^2+y^3

=x^3+y^3

j: (x+y)(x^2-xy+y^2)=x^3+y^3

k: (5x-2y)(x^2-xy-1)

=5x*x^2-5x*xy-5x-2y*x^2+2y*xy+2y

=5x^3-5x^2y-5x-2x^2y+2xy^2+2y

=5x^3-7x^2y+2xy^2-5x+2y

l: (x^2y^2-xy+y)(x-y)

=x^3y^2-x^2y^3-x^2y^2+xy^2+xy-y^2

$\begin{array}{c}xy-2x-3y=5\\ \Rightarrow xy-2x-3y+6=5+6\\ \Rightarrow x\left(y-2\right)-3\left(y-2\right)=11\\ \Rightarrow \left(x-3\right)\left(y-2\right)=11\end{array}$

Vì $11=1.11=\left(-1\right).\left(-11\right)$ nên ta có bảng sau:

Vậy $\left(x;y\right)=\left\{\left(-8;1\right);\left(2;-9\right);\left(4;13\right);\left(14;3\right)\right\}$.

(x+3y)^2 – x^2 + xy

= x^2+6xy + 9y^2-x^2 + xy

=9y^2 + 7xy = y( 9y+7x)

xy - 2x + 3y = 9

<=> x(y - 2) + 3y - 6 = 9 - 6

<=> x(y - 2) + 3(y - 2) = 3

<=> (x + 3)(y - 2) = 3

=> x + 3 và y - 2 là ước của 3

Ư(3) = { ± 1 ;± 3 }

Ta có bảng sau :

Vậy ( x;y ) = { ( - 6;1 ) ; ( - 4;-1 ) ; ( 0;3 ) ; ( - 2; 5 ) }

x(y-2)+3y-6=9-6

(x+3)(y-2)=3

\(x+3=\left(-3,-1,1,3\right)\)=> x=(-6,-4,-2,0)

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

KL (x,y)=(-6,1);(-4,-1);(-2,5);(0,3)