Bài 1:
a) Để \(\sqrt[]{9x^2-6x+1}\) xác định \(\Leftrightarrow9x^2-6x+1\ge0\Leftrightarrow\left(3x-1\right)^2\ge0\) ( luôn đúng )
Vậy \(\sqrt[]{9x^2-6x+1}\) luôn xác định với mọi giá trị của x
b) Để \(\sqrt[]{\dfrac{2x-1}{2-x}}\) xác định \(\Leftrightarrow\dfrac{2x-1}{2-x}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\2-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\2-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}\le x< 2\)
Bài 2:
a) \(\sqrt{\left(\dfrac{1}{\sqrt[]{2}}-\dfrac{1}{\sqrt{3}}\right)^2}=\left|\dfrac{1}{\sqrt[]{2}}-\dfrac{1}{\sqrt[]{3}}\right|=\dfrac{1}{\sqrt[]{2}}-\dfrac{1}{\sqrt[]{3}}\)
\(=\dfrac{\sqrt[]{2}}{2}-\dfrac{\sqrt[]{3}}{3}=\dfrac{3\sqrt[]{2}-2\sqrt[]{3}}{6}\)
b) Câu b có sai đề không vậy?
Bài 3: ( Sai đề ) Điều kiện: \(x\le3\)
\(\sqrt{\left(x-3\right)^2}=3-x\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow3-x\ge0\) \(\Leftrightarrow x\le3\)
Bài 2b:
\(\sqrt[]{\left(\sqrt[]{x}-\sqrt[]{y}\right)^2+\sqrt[]{x^2-8x+16}}\)
\(=\sqrt[]{x-2\sqrt[]{xy}+y+\sqrt[]{\left(x-4\right)^2}}\)
\(=\sqrt[]{x-2\sqrt[]{xy}+y+\left|x-4\right|}\)
\(=\left[{}\begin{matrix}\sqrt[]{x-2\sqrt[]{xy}+y+x-4}=\sqrt[]{2x-2\sqrt[]{xy}+y-4}\left(x\ge4\right)\\\sqrt[]{x-2\sqrt[]{xy}+y+4-x}=\sqrt[]{4-2\sqrt[]{xy}+y}\left(x< 4\right)\end{matrix}\right.\)
