a) Ta có: \(x^2\left(x-5\right)+x^2-4x-5=0\)
\(\Leftrightarrow x^2\left(x-5\right)+x^2-5x+x-5=0\)
\(\Leftrightarrow x^2\left(x-5\right)+x\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+x+1\right)=0\)(1)
Ta có: \(x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(2)
Từ (1) và (2) suy ra
x-5=0
\(\Leftrightarrow x=5\)
Vậy: x=5
b) Ta có: \(x^6-1=0\)
\(\Leftrightarrow x^6=1\)
\(\Leftrightarrow x=1\)
Vậy: x=1
