a*a+4a+1=0,且a*a*a*a+m*a*a+1/2a*a*a+ma*a+2a=3求m的值
问题描述：
a*a+4a+1=0,且a*a*a*a+m*a*a+1/2a*a*a+ma*a+2a=3求m的值
答
because(以下简写b): a*a+4a+1=0So, (a^2+2a+1)+2a=0 (a+1)^2+2a=0 [(a+1)^2+2(a+1)+1]-3=0 [(a+1)+1]^2-3=0 (a+2)^2=3 a+2=+/-根号3 a=+/-(根号3) -2 代入后面...