这题怎么解:(2x-1)^5=a+bx^2+cx^3+dx^4+ex^5 求:a+b+c+d+e=
(x+1)^5=ax^5+bx^4+cx^3+dX^2+ex+f,求a+b+c+d+e+f,b+c+d+e,a+c+e
已知(2x+1)^5=a+bx+cx^2+dx^3+ex^4+fx^5,求下列各式的值:(1)a+b+c+d+e+f;(
(2x-1)^5=ax^5+bx^4+cx^3+dx^2+ex+f 求a=?b=?c=?d=?e=?f=?
已知等式(2x-1)^5=ax^5+bx^4+cx^3+dx^2+ex+f,求代数式a+b+c+d+e+f的值.
(3x+1)^5=ax^5+bx^4+cx^3+dx^2+ex+f,求a-b+c-d+e-f
已知等式(3x+1)^5=ax^5+bx^4+cx^3+dx^2+ex+f,求代数式a-b+c-d+e-f的值
已知等式(3x+1)^5=ax^5+bx^4+cx^3+dx^2+ex+f,求代数式-a+b-c+d-e+f的值
(x+1)^5=ax^5+bx^4+cx^3+dx^2+ex+f 求b+c+d+e
(x-3)^5=ax^5+bx^4+cx^3+dx^2+ex+f ,则a+b+c+d+e+f= ,b+c+d+e= .
已知等式(x-3)^5=ax^5+bx^4+cx^3+dx^2+ex+f,求②a-b+c-d+e
(x+1)*6=x*6+ax*5+bx*4+cx*3+dx*2+ex+1 则a+b+c+d+e=?
若(2x+1)^5=ax^5+bx^4+cx^3+dx^2+ex+f,则a-b+c-d+e-f的值=