Question

Bài 6.Tìm số nguyên x sao cho: a) (x-1)^2=1 b)x.(3x+9)=0 c)52-(x+21)=19-x

Answer 1

a) Ta có: \(\left(x-1\right)^2=1\)

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

Vậy: \(x\in\left\{0;2\right\}\)

b) Ta có: x(3x+9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3\right\}\)