Question

a,x(16-x)=0
b,x(x+2)-x-2=0
c,\(x^2-9=0\)

Answer 1

a) x=0, x=16

b) x=-2, x=1

c) x=-3, x=3

 

Answer 2

a) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)

b) \(\Rightarrow x\left(x+2\right)-\left(x+2\right)=0\Rightarrow\left(x+2\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

c) \(\Rightarrow\left(x-3\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Answer 3

a)x=0 hoặc 16-x=0

x=0 hoặc x=16

Vậy \(x\in\left\{0;16\right\}\)

b)(x-1)(x+2)=0

x=1 hoặc x=-2

Vậy \(x\in\left\{1;-2\right\}\)

c)\(x^2=9\)

\(x=\pm3\)

Vậy \(x\in\left\{3;-3\right\}\)

Answer 4

a)\(x\left(16-x\right)=0\)

⇔\(x=0\) hay \(16-x=0\)

⇔\(x=0\) hay \(x=16\)

b)\(x\left(x+2\right)-x-2=0\)

⇔\(x\left(x+2\right)-\left(x+2\right)=0\)

⇔\(\left(x+2\right)\left(x-1\right)=0\)

⇔\(x+2=0\) hay \(x-1=0\)

⇔\(x=-2\) hay \(x=1\)

c)\(x^2-9=0\)

⇔\(\left(x-3\right)\left(x+3\right)=0\)

⇔\(x-3=0\) hay \(x+3=0\)

⇔\(x=3\) hay \(x=-3\)