(x-1) (x-2)(x-3)(x-4)(x-5)=0一元二次方程？
已知函数f(x)=x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5),则f'(0)=?
f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5) f'(x)=(x-1)(x-2)(x-3)(x-4)(x-5)+x(x-2).(x-5)+...+x(x-1)..(x-4) 只有第一大项不含因式x,其它5项均含因式x,将x=0代入时,后面均是0 f'(0)=(-1)(-2)(-3)(-4)(-5)=-120
计算(x - 1)(x - 2)(x - 3)(x - 4).
(x-1)(x-2)(x-3)(x-4)=(x-1)(x-4)(x-2)(x-3)=(x^2-5x+4)(x^2-5x+6)=(x^2-5x)^2+10(x^2-5x)+24=x^4-10x^3+25x^2+10x^2-50x+24=x^4-10x^3+35x^2-50x+24
(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)的展开式中,x的四次方的系数是多少?
-1-2-3-4-5=-15
(X - 1)(X - 2)(X - 3)(X - 4)=?简便方法
(X-1)(X-2)(X-3)(X-4) = [(x -1)(x-4)] * [(x-2)(x-3)] = [x^2 - 5x +4] * [[x^2 - 5x + 6] = (x^2 - 5x)^2 + (4 + 6)* (x^2 - 5x) + 4 * 6 = (x^2 - 5x)^2 + 10(x^2 - 5x) + 24 = x^2 - 10x^3 + 25x^2 + 10x^2 - 50x + 24 = x^4 - 10x^3 + 35x^2 - 50x + 24
(x+1)(x - 2)(x - 3)/(x - 4)(x - 5)>0的解集是
x+1>x-2>x-3>x-4>x-5+ + + / + + √ ==>x-5>0 ==>x>5+ + + / + - X+ + + / - - √ ==>x-3>0 且 x-4<0 ==>3<x<4+ + - / - - X+ - - / - - X- - - / - - √ ==>x+1<0 ==>x<-1x∈(-∞,-1)∪(3,4)∪(5,+∞)
在(X - 1)(X - 2)(X - 3)(X - 4)(X - 5)的展开式中,含X的3次方的项的系数是
1*2+1*3+1*4+1*5+2*3+2*4+2*5+3*4+3*5+4*5=14+6+8+10+27+20=85
在(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)的展开式中,含x^4的项的系数是
(x-1)(x-2)(x-3)(x-4)(x-5)的展开式中,含x^4的项的系数是-1-2-3-4-5=-15
f(x)=(x - 1)(x - 2)(x - 3)(x - 4).问方程f`(x)=0有几个实根,并指出它们所在.
(x-1)(x-2)(x-3)(x-4)=0x-1=0或x-2=0或x-3=0或x-4=0解得x=1或x=2或x=3或x=4
设函数f(x)=(x - 1)(x - 2)(x - 3)(x - 4)(x - 5),则f'(3)=
函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)=(x-3)[(x-1)(x-2)(x-4)(x-5)] f(x) '=[(x-1)(x-2)(x-4)(x-5)]+(x-3)[(x-1)(x-2)(x-4)(x-5)]' f(3) '=(3-1)x(3-2)(3-4)(3-5)+0=2x1x1x2=4
(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)=720 求过程! 过程详细采纳!
成绩差赶紧去补补啦