a/ ĐKXĐ: \(\sqrt{x^2-4x+4}\le x\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}\le x\) \(\Leftrightarrow\sqrt{\left(2-x\right)^2}\le x\Leftrightarrow2-x\le x\Leftrightarrow x\ge1\)
b/ \(A=\sqrt{x-\sqrt{x^2-4x+4}}=\sqrt{x-\sqrt{\left(x-2\right)^2}}\)
\(=\sqrt{x-\left(x-2\right)}=\sqrt{x-x+2}=\sqrt{2}\)
c/ Thay x = \(3-\sqrt{3}\) vào A ta có:
\(A=\sqrt{3-\sqrt{3}-\sqrt{\left(3-\sqrt{3}\right)^2-4\cdot\left(3-\sqrt{3}\right)+4}}\)
\(=\sqrt{3-\sqrt{3}-\sqrt{\left(3-\sqrt{3}-2\right)^2}}\)
\(=\sqrt{3-\sqrt{3}-\left(3-\sqrt{3}-2\right)}\)
\(=\sqrt{3-\sqrt{3}-3+\sqrt{3}+2}=\sqrt{2}\)
a)A=\(\sqrt{x-\sqrt{\left(x+2\right)^2}}=\sqrt{x-\left|x-2\right|}\)
ĐKXĐ của A là :
x\(x\ge\left|x-2\right|\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2\ge x^2-4x+4\end{matrix}\right.\Leftrightarrow x\ge1\)
b) Nếu \(x\ge2\) thì A=\(\sqrt{x-\left(x-2\right)}=\sqrt{2}\)
Nếu 1\(\le x\)<2 thì A =\(\sqrt{x-\left(x-2\right)}=\sqrt{2x-2}\)
