x2-2ax 2
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/19 09:02:56
![x2-2ax 2](/uploads/image/f/891632-56-2.jpg?t=x2-2ax+2)
显然a≠0由△=16a2-16a(a+4)≥0得a<0,由韦达定理知x1+x2=1,x1x2=a+44a,所以(x1−2x2)(x2−2x1)=5x1x2−2(x21+x22)=9x1x2−2(x1+
与x轴的两个交点就是(-2,0)与(1,0)再问:给讲下不再答:y=ax^2+bx+c当x=x1时,y=0,因此(x1,0)为曲线上一个点,此点在x轴上。当x=x2时,y=0,因此(x2,0)为曲线上
f(x1)-f(x2)=ax1^2+2ax1+4-ax2^2-2ax2-4=a(x1^2-x2^2)+2a(x1-x2)=a(x1-x2)(x1+x2+2)=a(x1-x2)(3-a)因为X1<X2、
f(2分之x1+x2)与f(x1+x2)都等于最值.即都等于(4ac-b^2)/4a.再问:还有f(x1+x2)等于多少?再答:X1+X2=-b/af(X1+X2)=a*b^2/(a^2)+b*(-b
2a(x2+1)2-2ax2,=2a[(x2+1)2-x2],=2a(x2+x+1)(x2-x+1).故答案为:2a(x2+x+1)(x2-x+1).
设x4+ax2+b=(x2+2x+5)(x2+mx+n)=x4+(2+m)x3+(2m+n+5)x2+(5m+2n)x+5n比较对应项系数得2+m=02m+n+5=a5m+2n=05n=b解得m=-2
由于x2+3x+2=(x+1)(x+2),假如f(x)能被x2+3x+2整除,则(x+1)和(x+2)必是f(x)的因式,因此,当x=-1时,f(-1)=0,即1+a+b+2=0,①当x=-2时,f(
(I)f'(x)=3x2+4ax+b,g'(x)=2x-3.由于曲线y=f(x)与y=g(x)在点(2,0)处有相同的切线l.故有f(2)=g(2)=0,f'(2)=g'(2)=1.由此得{8+8a+
(2ax2-x2+3x+2)-(5ax2-4x2+3x)=2ax2-x2+3x+2-5ax2+4x2-3x=(-3a+3)x2+2,∵多项式(2ax2-x2+3x+2)-(5ax2-4x2+3x)的值
y=-(1/2)(x-2)(x-3)=-(1/2)x^2+(5/2)x-3
由于(ax2-2xy+y2)-(-x2+bxy+2y2)=(a+1)x2-(2+b)xy-y2=5x2-9xy+cy2;若令上面等式成立,需满足a+1=52+b=9c=−1;解得:a=4b=7c=−1
已知等式变形得:2ax2-2x3+bx2+abx=-2x3+(2a+b)x2+abx=ax3-3x2-cx,则a=-2,2a+b=-3,ab=-c,解得:a=-2,b=1,c=2,则a+b+c=-2+
1.选A.因为对称轴是x=-1,抛物线开口向上,点x1与x2关于原点对称.结合图象可知,选A.2.选D填空题:1.这是以及-1为圆心,以2为半径,在x轴上方的半圆,所以单调递增区域是[-3,-1]2.
①A+B=ax2+x-1+3x2-2x+1=(a+3)x2-x∵A与B的和中不含x2项,∴a+3=0,解得a=-3.②B-2A=3x2-2x+1-2×(-3x2+x-1)=3x2-2x+1+6x2-2
∵等式的左边=3x2-3x+2+x2-3x+3=(3+1)x2-(3+3)x+2+3=4x2-6x+5,∴A=4,B=6,C=5.故选D.
∵x2+x-2=(x+2)(x-1),∴2x4-3x3+ax2+7x+b能被(x+2)(x-1)整除,设商是A.则2x4-3x3+ax2+7x+b=A(x+2)(x-1),则x=-2和x=1时,右边都
11-1123a31a32r2-2r1,r3-r111-1101a+210a-141r3-(a-1)r211-1101a+2100-(a-2)(a+3)-(a-2)当a=-3时,无解当a=2时,无穷多
(1)f'(x)=-2x^2+2x+2在0≤x≤1上大于0故递增得0
选af(x)=ax2+2ax+4=a(x+1)^2-a+4a>0,所以图象是开口向上,x=-1为对称轴又x1+x2=0,所以x1与x2关于原点对称.由图象很容易看出f(x1)<f(x2)
x3+ax2-2x2+3x-bx-c=x3+(a-2)x2+(3-b)x-c根据题意得,a-2=-33-b=4-c=-1所以a=-1b=-1c=1