Question

急急急！在线等！一道数学题！求详解！多谢啦！

已知f(x)是二次函数，且f(2x)＋f(3x＋1)=13x^2＋6x－1 求f(x)

Answer 1

【参考答案】 f(x)=x²-1

令f(x)=ax²+bx+c
∵f(2x)＋f(3x＋1)=13x^2＋6x－1
∴a(2x)²+b×2x+c+a(3x+1)²+b(3x+1)+c=13x²+6x-1
4ax²+2bx+c+9ax²+6ax+a+3bx+b+c=13x²+6x-1
(13a-13)x²+(6a+5b-6)x+(a+b+2c+1)=0
∴ 13a-13=0
6a+5b-6=0
a+b+2c+1=0
解得 a=1, b=0, c=-1
∴ f(x)=x²-1

Answer 2

设F(x)=Ax^2+Bx+C
F(2x)=4Ax­­^2+2Bx+C
F(3x+1)=A(3x+1)^2+B(3x+1)+C
=A(9x^2+6x+1)+3Bx+B+C
=9Ax^2+6Ax+A+3Bx+B+C
=9Ax^2+(6A+3B)x+A+B+C
F(2x)+ F(3x+1)=4Ax­­^2+2Bx+C+ 9Ax^2+(6A+3B)x+A+B+C
=13Ax^2+(6A+5B)x+A+B+2c

对比各项序数得
A=1
B=0
C=-1
即得F(x)=x^2-1

Answer 3

设f(x）=ax^2+bx+c,a不等于0，代入式子里，就会有三个方程，可以依次得到：a,b,c的值，即可得f(x)

Answer 4

设f(x) = ax^2 + bx + c (a ≠ 0)f(2x) = 4ax^2 + 2bx + c f(3x+1) = a(3x+1)^2 + b(3x+1) + c = 9ax^2 + (6a+3b)x + a + b + 2cf(2x) + f(3x+1) = 13ax^2 + (6a+5b)x + a + b + c = 13x^2 + 6x -1比较系数：a = 16a+5b = 6 +5b =6，　b= 0a + b + 2c=1+0+ 2c= 1+ 2c = -1c = -1f(x) = x^2 -1

Answer 5

既然知道是二次函数很简单设f(x) = ax^2 + bx + c (a ≠ 0)然后代入就可以了