\(x^2(x-5)+x^2-4x-5=0\)
\(⇔x^2(x-5)+x^2+x-5x-5=0 \)
\(⇔x^2(x-5)+(x+1)(x-5)=0 \)
\(⇔(x-5)(x^2+x+1)=0\)
\(⇔\left[\begin{array}{} x-5=0\\ x^2+x+1=0(mà:x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}>0) \end{array}\right.\)
\(⇔x=5\)
Vậy pt có 1 nghiệm là x=5