Question

数学思考题！急！已知abct≠0，满足a=bt+c，b=c（1+t+t^2）.（1）求证：a是二次方程cx^2+c（b-

Answer 1

a=c（1+t+t^2)t+c=c(t+1)(t^2+1) 证明：把a代入c^3*(t+1)^2*(t^2+1)^2+c^3*(t^2+t-1)*(t+1)(t^2+1)-c^3*(t^2+t)((1+t+t^2)^2+1)=c^3[(t+1)(t^2+1)(t^3+2t^2+2t)-(t^2+t)((1+t+t^2)^2+1)]=c^3[t(t+1)(t^4+2t^3+3t^2+2t+2-t^4-2t^3-3t^2-2t-2)=0 把a=15，b=7代入得到c^3-37c^2+379c-343=0(c-1)(c^2-36c+343)=0c=1t=2