maple 如何化简assume(a > 0,b > 0,c > 0,m > n,n > 0,x > 1/2,x < 1);factor(-c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2*x*n));u := %;u := -c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2
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maple 如何化简assume(a > 0,b > 0,c > 0,m > n,n > 0,x > 1/2,x < 1);factor(-c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2*x*n));u := %;u := -c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2
maple 如何化简
assume(a > 0,b > 0,c > 0,m > n,n > 0,x > 1/2,x < 1);
factor(-c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2*x*n));
u := %;
u := -c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2*x*n)
type(u,polynom);
false
想化简一个有理式,分子分母约去公因式,这个u是分子,已经做了假设,为什么还是判断u不是多项式?此外还有什么办法能把有理式化成我想要的形式么?
maple 如何化简assume(a > 0,b > 0,c > 0,m > n,n > 0,x > 1/2,x < 1);factor(-c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2*x*n));u := %;u := -c*m^2+c*m*n+2*b*n(-2*x^2*n+m(1-2*x)^2+2*x*n)+2*c*m-4*a(-2*x^2*n+m(1-2*x)^2+2
式中丢了三处乘号,这样 Maple 会不知所措,电脑毕竟是机器,软件也非万能；
无需做上述假设,即可判断它是多项式；
其他可用的命令有 factor、normal、expand 等
过来瞅瞅
simplify()命令试试