Question

Bài 1: Tìm x \(\in\) Z, biết :

a) x.(x + 2) = 0

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

c) (x - 2)(\(x^2\) + 1) = 0

d) (x + 1)(\(x^2\) - 4) = 0

Answer 1

a) \(x.\left(x+2\right)=0\)

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

Vậy \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b) \(\left(x-1\right)\left(x-2\right)=0\)

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

Vậy \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

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

Vậy \(x=2\)

d) \(\left(x+1\right)\left(x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-4=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

Answer 2

A) x={0;-2}

B) x={1;2}

C) x=2

D) x={-2;-1;2}