Question

Giải phương trình\(\sqrt{x^2-4x+4} = \sqrt{7-4 \sqrt{3}}\)

Answer 1

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(\Leftrightarrow\left|x-2\right|=2-\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2-\sqrt{3}\\x-2=\sqrt{3}-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)

PT có tập nghiệm : \(S=\left\{4-\sqrt{3};\sqrt{3}\right\}\)

Answer 2

\(\sqrt{x^2-4x+4}=\sqrt{7-4\sqrt{3}}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(\Leftrightarrow\left|x-2\right|=2-\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2-\sqrt{3}\\-x+2=2-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)

Vậy \(x=\sqrt{3}\) hoặc \(x=4-\sqrt{3}\)

Answer 3

\(\sqrt{x^2-4x+4}=\sqrt{7-4\sqrt{3}}\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{4-4\sqrt{3}+3}\Leftrightarrow\left|x-2\right|=\sqrt{\left(2-\sqrt{3}\right)^2}\Leftrightarrow\left|x-2\right|=\left|2-\sqrt{3}\right|\Leftrightarrow\left|x-2\right|=2-\sqrt{3}\)\(\Leftrightarrow\left[{}\begin{matrix}x-2=2-\sqrt{3}\\x-2=\sqrt{3}-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4-\sqrt{3}\left(tm\right)\\x=\sqrt{3}\left(tm\right)\end{matrix}\right.\)

Vậy S={\(\sqrt{3};4-\sqrt{3}\)}

Answer 4

\(\sqrt{\left(x-2\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^3}\\ x-2=2-\sqrt{3}\\ x-4+\sqrt{3}=0\\ x=4-\sqrt{3}\)