(x-1)(x-2)

薄萱炕2909y=(x - 1)(x - 2)(x - 3)(x - 4)的导数 -
殷秋贸17747269603 ______[答案] y=(x-1)(x-2)(x-3)(x-4)=(x^2-5x)^2+10(x^2-5x)+24 y'=2(x^2-5x)(2x-5)+10(2x-5) =(2x-5)(2x^2-10x+10)

薄萱炕2909分解因式:(x - 1)(x - 2)(x - 3)(x - 4) - 48 -
殷秋贸17747269603 ______ (x-1)(x-2)(x-3)(x-4)-48=(x-1)(x-4)(x-2)(x-3)-48=(x^2-5x+4)(x^2-5x+6)-48=(x^2-5x)^2+10(x^2-5x+6)-24=(x^2-5x-2)(x^2-5x+12)

薄萱炕2909分解因式:(x - 1)(x - 2)(x - 3)(x - 4) - 24 -
殷秋贸17747269603 ______[答案] (x-1)(x-2)(x-3)(x-4)-24 =(x-1)(x-4)(x-2)(x-3)-24 =(x^2-5x+4)(x^2-5x+6)-24 =(x^2-5x)^2+10(x^2-5x) +24-24 =(x^2-5x)(x^2-5x+10) =x(x-5)(x^2-5x+10)

薄萱炕2909解方程:(x - 1)(x - 2)(x - 3)(x - 4)=1*2*3*4 -
殷秋贸17747269603 ______[答案] (x-1)(x-4)=(x^2-5x)+4 (x-2)(x-3)=(x^2-5x)+6 所以,(x-1)(x-2)(x-3))(x-4)=(x^2-5x)^2+10(x^2-5x)+24 方程(x-1)(x-2)(x-3)(x-4)=1*2*3*4化为(x^2-5x)^2+10(x^2-5x)=0 得x^2-5x=0或x^2-5x+10=0,解得x=0或5 所以,方程有二0和5

薄萱炕2909fx=(x - 1)(x - 2)……(x - 2014)求f(2014)的导 -
殷秋贸17747269603 ______ 解由f'x=[(x-1)(x-2)……(x-2014)]'=[(x-1)(x-2)……(x-213)]'(x-2104)+(x-2014)'(x-1)(x-2)……(x-2013)=[(x-1)(x-2)……(x-213)]'(x-2104)+(x-1)(x-2)……(x-213) 故f(2014)=[(2014-1)(2014-2)……(2014-213)]'(2014-2104)+(2014-1)(2014-2)……(2014-2013)=0+2013!

薄萱炕2909求教x/(x - 1)(x - 2)的n阶导数 -
殷秋贸17747269603 ______ 原题是:求x/((x-1)(x-2))的n阶导数. 设y=x/((x-1)(x-2)) y=2/(x-1)-1/(x-2) y'=-2/(x-2)^2+1/(x-2)^2 y''=2*2/(x-2)^3-2/(x-2)^3=(-1)^2*2!(2/(x-2)^3-1/(x-1)^3) y'''=-2*3!/(x-2)^4+3!/(x-2)^4=(-1)^3*3!(2/(x-2)^4-1/(x-1)^4) ... y(n)=(-1)^n*n!(2/(x-2)^(n+1)-1/(x-1)^(n+1))

薄萱炕2909(x - 1)(x - 2)(x - 3)(x - 4) - 120 化简的, -
殷秋贸17747269603 ______[答案] (x-1)(x-2)(x-3)(x-4)-120=[(X-1)(X-4)][(X-2)(X-3)]-120=[(X²-5X)+4][(X²-5X)+6]-120=(X²-5X)²+10(X²-5X)+24-120=(X²-5X)²+10(X²-5X)-96=(X²-5X-6)(X²-5X+16)=...

薄萱炕2909(x - 1)(x+2)=0 怎么算. -
殷秋贸17747269603 ______[答案] (x-1)(x+2)=0 (x-1)=0 x=1 x+2=0 x=-2

薄萱炕2909化简丨x - 1丨+丨x - 2丨 -
殷秋贸17747269603 ______ 当x丨x-1丨+丨x-2丨 =1-x+2-x =3-2x 当1≤x≤2时 丨x-1丨+丨x-2丨 =x-1+2-x =1 当x>2时 丨x-1丨+丨x-2丨 =x-1+x-2 =2x-3

薄萱炕2909求导y=根号下(2x+3)(x - 1)/(2x - 1)(x+2) -
殷秋贸17747269603 ______[答案] y=√{(2x+3)(x-1)/[(2x-1)(x+2)]} =√[(2x+3)(x-1)]/√[(2x-1)(x+2)]=u/v y'=(u/v)'=[u'v-uv']/v^2 ={1/2*1/√[(2x+3)(x-1)]*[2*(x-1)+(2x+3)*1]*√[(2x-1)(x+2)]-√[(2x+3)(x-1)]*1/2*1/√[(2x-1)(x+2)]*[2*(x+2)+(2x-1)*1]}/{√[(2x-1)(x+2)]}^2 =(4x+1)]*√[(2x-1)(x+2)]/√[(2x+3)(x-...