a) \(x^2\left(x-5\right)+x^2-4x-5=0\)
⇔\(x^2\left(x-5\right)+\left(x-5\right)\left(x+1\right)=0\)
⇔\(\left(x-5\right)\left(x^2+x+1\right)=0\)
Vì \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)∀x
⇒\(x=5\)
Vậy ...
b) \(x^8-1=0\)
⇔\(x^8=1\)
⇒ \(x=+-1\)