Question

Answer 1

a, \(x^2\left(x-5\right)+\left(x-2\right)^2-9=0\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=5\)

b, \(x^6-1=0\Leftrightarrow x^6=1\Leftrightarrow x=\pm1\)

Answer 2

a: \(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)\left(x+1\right)=0\)

=>x-5=0

hay x=5

b: \(\Leftrightarrow x^2-1=0\)

hay \(x\in\left\{1;-1\right\}\)

Answer 3

\(a,PT\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)\left(x+1\right)=0\Leftrightarrow\left(x-5\right)\left(x^2+x+1\right)\Leftrightarrow x=5\) (Vì \(x^2+x+1>0\forall x\))

Vậy: \(S=\left\{5\right\}\)

\(b,PT\Leftrightarrow x^6=1\Leftrightarrow x=\pm1\)

Vậy: \(S=\left\{\pm1\right\}\)