MT【102】一个常见的因式分解公式
解答:
$x^3+y^3+1-3xy=(x+y+1)(x^2+y^2+1+xy-x-y)=$
$(x+y+1)(x^2+y^2+1+xy-x-y)=$
$\frac{1}{2}(x+y+1)[(x-y)^2+(x-1)^2+(y-1)^2]\therefore x=y=1$
懂，会，熟，巧；趁青春尚在，奋力前行，追求卓越！
解答:
$x^3+y^3+1-3xy=(x+y+1)(x^2+y^2+1+xy-x-y)=$
$(x+y+1)(x^2+y^2+1+xy-x-y)=$
$\frac{1}{2}(x+y+1)[(x-y)^2+(x-1)^2+(y-1)^2]\therefore x=y=1$
