\(x^2=x^5\)
\(\Rightarrow x^2-x^5=0\)
\(\Rightarrow x^2.\left(1-x^3\right)=0\)
+) \(x^2=0\Rightarrow x=0\)
+) \(1-x^3=0\)
\(\Rightarrow x^3=1\)
\(\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
x2 = x5
→ x2 - x5 = 0
→ 22. ( 1 - x3 ) = 0
+) x2 = 0 → x = 0
+) 1 - x3 = 0
→ x3 = 1
→ x = 1
Vậy x ϵ { 0 ; 1 }