f(x - 1) = x² - 2,f(x) = (x + 1)² + 1f(2x) = 2x + 1,f(x) = 2×(1/2)x + 1 = x + 1f(x) = x + 4f(ax + b) = ax + b + 4ax + b + 4 = x + 10(a - 1)x + (b + 4 - 10) = 0a - 1 = 0,a = 1b + 4 - 10 = 0b...1.已知函數(shù)f(x)=x平方+2x+a,f(bx)=9乘x平方-6x+2，其中x屬于R,a,b為實常數(shù)，求方程f(ax+b)=0的解2.已知二次函數(shù)y=f(x)滿足f(x+1)=f(1-x)(1)求證x=1是y=f(x)圖像的對稱軸 (2)若二次函數(shù)y=f（x）圖像又過兩點（1,1）和（0,3），求其解析式“搭的比賣的還多”，行！f(x) = x² + 2x + af(bx) = b²x² + 2bx + a9x² - 6x + 2 = b²x² + 2bx + aa = 2b = ±3f(ax + b) = f(2x ± 3) = (2x ± 3)² + 2(2x ± 3) + 2 = 0(2x ± 3 + 1)² + 1 = 0無解