Đáp án:a. x+xy-y
b.(3;0) (-1;-2) (2;1) (0;-3)
Giải thích các bước giải:
a. P=$\frac{x^2(1+x)-y^2(1-y)-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$
=$\frac{x^3+x^2-y^2+y^3-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$
=$\frac{(x^3+y^3)+(x^2-y^2)-x^2y^2(x+y)}{(x+y0(1+x)(1-y)}$
=$\frac{(x+y)(x^2-xy+y^2)+(x-y)(x+y)-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$
=$\frac{(x+y)(x^2-xy+y^2+x-y-x^2y^2)}{(x+y)(1+x)(1-y)}$
=$\frac{x^2-xy+y^2+x-y-x^2y^2}{(1+x)(1-y)}$
=$\frac{(x^2+x)-(xy+y)+(y^2-x^2y^2)}{(1+x)(1-y)}$
=$\frac{x(1+x)-y(x+1)+y^2(1-x)(1+x)}{(1+x)(1-y)}$
=$\frac{(x+1)(x-y-xy^2+y^2)}{(x+1)(1-y)}$
=$\frac{x-y-xy^2+y^2}{1-y}$
=$\frac{x(1-y^2)-y(1-y)}{1-y}$
=$\frac{x(1-y)(1+y)-y(1-y)}{1-y}$
=x+xy-y
b.
P=3 ⇔ x+xy-y=3
⇔x(1+y)=y+3
⇔x=$\frac{y+3}{y+1}$ ⇔x=1+$\frac{2}{y+1}$
do x ∈Z ⇒ 1+$\frac{2}{y+1}$ ∈Z
⇒ 2 chia hết cho (y+1)
⇒y+1 ∈ Ư(2)={ ±1;±2}
y+1=1⇒y=0; x=3
y+1=-1⇒y=-2;x=-1
y+1=2⇒y=1; x=2
y+1=-2⇒y=-3;x=0
