Question

(x-2)^2=1

Answer 1

Ta có: \(\left(x-2\right)^2=1\)

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\).

Answer 2

\(\left(x-2\right)^2=1\\ \Leftrightarrow x-2=\pm1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy \(x_1=1;x_2=3\)

Answer 3

\(\left(x-2\right)^2=1\) \(\Leftrightarrow\) \(\sqrt{\left(x-2\right)^2}=\sqrt{1}\) \(\Leftrightarrow\) \(\left|x-2\right|=1\)

trường hợp 1 : \(x\ge2\)

\(\Rightarrow\) \(\left|x-2\right|=1\Leftrightarrow\) \(x-2=1\) \(\Leftrightarrow\) \(x=3\) (tmđk)

trường hợp 2 : \(x< 2\)

\(\Rightarrow\) \(\left|x-2\right|=1\) \(\Leftrightarrow\) \(2-x=1\) \(\Leftrightarrow\) \(-x=-1\) \(\Leftrightarrow\) \(x=1\) (tmđk)

vậy \(x=3;x=1\)

Answer 4

\(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left(x-2\right)^2=\pm1^2\)

\(\Leftrightarrow x-2=1\Leftrightarrow x=3\)

\(\Leftrightarrow x-2=-1\Leftrightarrow x=1\)

Answer 5

\(\left(x-2\right)^2=1\)

\(\Rightarrow\left(x-2\right)^2=\left(\pm1\right)^2\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)