ax^2 (2a 1)x 1>0 -1
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ax²-(3a+1)x+2(a+1)=0a-(a+1)1-2(ax-(a+1))(x-2)=0x=2x=(a+1)/a=1+1/aa≠0两个不相等的实数根(a+1)/a≠2a+1≠2aa≠1
f(x)=ax^2+2ax+4=a(x+1)^2-a+4∵x10∴f(x1)-f(x2)=[a(-x2+1)^2-a+4]-[a(x2+1)^2-a+4]=a(-x2+1+x2+1)(-x2+1-x2
X1,X2是方程X^2-2aX+a+b=0两实数根x1+x2=2ax1*x2=a+b且△=(-2a)^2-4(a+b)≥0a^2≥a+b=x1*x2(X1-1)^2+(X2-1)^2=(x1^2-2x
1)原式=√[(x1-x2)²]=√[(x1+x2)²-4x1x2]因为x1+x2=-b/a,x1x2=c/a所以原式=√(b²/a²-4c/a)2)原式=(x
给你个思路,显然有a1,……an线性无关(由范德蒙德行列式不为0容易证明)因此得证我先回答的>_
对称轴x=-1(x1+x2)/2=(1-a)/20
由题意,f(x)有三个解,可必可以分解因式,即f(x)=x(x-1)(x-2)=x^3-3x^2+2xf'(x)=3x^2-6x+2令f'(x)=0,即3x^2-6x+2=0设两根为x1,x2,由韦达
x1+x2=1-a--->x1=1-a-x2f(x1)-f(x2)=a(x1²-x2²)+2a(x1-x2)+(4-4)=a(x1+x2)(x1-x2)+2a(x1-x2)=[a(
(x1)+(a1)(x2)+(a1)^2(x3)=1(1)(x1)+(a2)(x2)+(a2)^2(x3)=1(2)(x1)+(a3)(x2)+(a3)^2(x3)=1(3)(2)-(1)得(a2-a
f(x)=ax^2+2ax+4=a(x+1)^2-a+4因为x10所以f(x1)-f(x2)=[a(-x2+1)^2-a+4]-[a(x2+1)^2-a+4]=a(-x2+1+x2+1)(-x2+1-
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)+2a(x1-x2)=a(x1-x2)(x1+x
因为x1x2=c/a,x1+x2=-b/a(其中,a=1,b=-a,c=a^2-a+(1/4)),则,x1x2/(x1+x2)=a-1+(1/4a)∵Δ=a²-4(a²-a+1/4
因为x1,x2是关于x方程x^2-ax+a^2-a+(1/4)=0的两个实根,所以(1)△≥0,即a^2-4a^2+4a-1≥0,从而1≥a≥1/3(2)(x1x2)/(x1+x2)=a+1/4a-1
y²/4+x²=1【是椭圆!】直线PA₁:(y-2)/x=(y₁-2)/x₁直线PA₂:(y+2)/x=(y₁+2)/(
即[f(x1)+x1-f(x2)+x2]/(x1-x2)>0所以令g(x)=f(x)+xg'(x)=x-(a-1)+(a-1)/x=[x^2-(a-1)x+a-1]/a1
这道题算是比较典型的吧第一题af(-1)再问:f(-2)f(0)
(1)x^2-ax+2=0x10解集{x|x>0}(3)log(1/2)x
关于X的方程f(x)=ax^2-2bx+2-b=0(a>0)的两根X1、X2满足0再问:就是以a,b为x,y轴吧谢了