M=$\frac{x^{{2}}}{(x+y)(1-y)}$ -$\frac{y^{{2}}}{(x+y)(1+x)}$ -$\frac{x^{{2}}y^{{2}}}{(1+x)(1-y)}$
M= $\frac{x^{{2}}(x+1)}{(x+y)(1-y)(x+1)}$-$\frac{y^{{2}}(1-y)}{(x+y)(1+x)(1-y)}$-$\frac{x^{{2}}y^{{2}}(x+y)}{(1+x)(1-y)(x+y)}$
M=$\frac{x^{{2}}(x+1)-y^{{2}}(1-y)-x^{{2}}y^{{2}}(x+y)}{(x+y)(1-y)(x+1)}$
M=$\frac{x^{3}+x^{2}-y^{2}+y3-x^{{2}}y^{{2}}(x+y)}{(x+y)(1-y)(x+1)}$
M=$\frac{x^{3}+y^{3}+x^{2}-y^{2}-x^{{2}}y^{{2}}(x+y)}{(x+y)(1-y)(x+1)}$
M= $\frac{(x+y)(x^{2}-xy-y^{2})-(x-y)(x+y)-x^{{2}}y^{{2}}(x+y)}{(x+y)(1-y)(x+1)}$
M=$\frac{(x+y)(x^{2}-xy+y^{2}+x-y-x^{{2}}y^{{2}})}{(x+y)(1-y)(x+1)}$
M=$\frac{(x+1)(1-y)(x+xy-y)}{(1-y)(x+1)}$
M=x+xy-y
b).M=x+xy-y=7
x+xy-y=7
x(1+y)-y-1=7-1
(1+y)(x-1)=6
Th1:1+y=6⇒y=5
x-1=1⇒x=2
tương tự 7 th còn lại
Vậy (x;y) là : (2;5);(0;-7);(7;0);(-5;0);(3;2);(-1;-4);(4;1);(-2;-3)
Giải thích các bước giải:
