已知函数f(x)解析式为2x²+x,解不等式[f(x)]²+2x²+x

已知函数f(x)解析式为2x²+x,解不等式[f(x)]²+2x²+x<0
RT

[f(x)]²+2x+x＜0
[f(x)]²+f(x)＜0
f(x)(f(x)+1)＜0
-1＜f(x)＜0
-1＜2x²+x＜0
2x²+x+1＞0①
x(2x+1)＜0②
①x∈R②-1/2＜x＜0
由①、②得解集{x|-1/2＜x＜0}.