a) Ta có: VT=\({\left( {\sqrt 3 - 1} \right)^2} = {\left( {\sqrt 3 } \right)^2} - 2. \sqrt 3 .1 + {1^2}\)
\( = 3 - 2\sqrt 3 + 1\)
\(=(3+1)-2\sqrt 3 \)
\(= 4 - 2\sqrt 3 \) = VP
Vậy \((\sqrt{3}- 1)^{2}= 4 - 2\sqrt{3}\) (đpcm)
b) Ta có:
VT=\(\sqrt {4 - 2\sqrt 3 } - \sqrt 3 = \sqrt {\left( {3 + 1} \right) - 2\sqrt 3 } - \sqrt 3 \)
\( = \sqrt {3 - 2\sqrt 3 + 1} - \sqrt 3 \)
\(= \sqrt {{{\left( {\sqrt 3 } \right)}^2} - 2.\sqrt 3 .1 + {1^2}} - \sqrt 3 \)
\( = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - \sqrt 3 \)
\( = \left| {\sqrt 3 - 1} \right| - \sqrt 3 \).
Lại có:
\(\left\{ \matrix{ {\left( {\sqrt 3 } \right)^2} = 3 \hfill \cr {\left( {\sqrt 1 } \right)^2} = 1 \hfill \cr} \right.\)
Mà \(3>1 \Leftrightarrow \sqrt 3 > \sqrt 1 \Leftrightarrow \sqrt 3 > 1 \Leftrightarrow \sqrt 3 -1 > 0 \).
\(\Rightarrow \left| \sqrt 3 -1 \right| = \sqrt 3 -1\).
Do đó \(\left| {\sqrt 3 - 1} \right| - \sqrt 3 = \sqrt 3 -1 - \sqrt 3\)
\(= (\sqrt 3 - \sqrt 3) -1= -1\) = VP.
Vậy \(\sqrt {4 - 2\sqrt 3 } - \sqrt 3 =-1\) (đpcm)
