f1(x) +f2(x)
＝（a1+a2)x^2+(b1+b2)x+c1+c2
实数R上的递增,则有：
a1+a2=0,b1+b2>0
所求的条件是：
a1+a2=0,b1+b2>0
若二次函数f1（x）=ax2+bx+c和f2（x）=ax+bx+c,使得
若二次函数f1（x）=ax2+bx+c和f2（x）=ax+bx+c,使得
若二次函数f1(x)=a1x2+b1x+c1和f2(x)= a2x2+b2x+c2,使得f1(x) +f2(x)实数R上的递增函数的条件是
若二次函数f1(x)=a1x2+b1x+c1和f2(x)= a2x2+b2x+c2,使得f1(x) +f2(x)实数R上的递增函数的条件是
数学人气：729 ℃时间：2020-04-02 21:11:24
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