设 f(x) = k1 x - 2 ,g(x) = k2 x则由 f(g(x)) = k1 k2 x - 2 = 3x - 2 , g(f(x)) = k2 ( k1 x - 2 ) = 3x - 2得出 k1 = 3, k2 = 1故袭扮则直线 y=f(x) = 3x-2 , y = g(x) = x 两直线的交点即为次缺羡方程拍棚组的解 x=1,y=1
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