Question

1.Tìm x

a/ x^3 - 16x = 0

b/x^2 - 6x + 9 = 0

 

Answer 1

a) \(x^3-16x=0\)

\(\Leftrightarrow x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{16}=\pm4\end{cases}}\)

Vậy \(x\in\left\{0;\pm4\right\}\)

Answer 2

b) \(x^2-6x+9=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Answer 3

\(a)x^3-16x=0=>x\left(x^2-16\right)=0\)

\(=>x\left(x-4\right)\left(x+4\right)=0\)

Suy ra x=0 hoặc x-4=0 hoặc x+4=0

=> x=0 hoặc x=4 hoặc x=-4

b)\(x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)