Bài 1:
a) \(\sqrt {64} = 8\)
b) \(\sqrt {25} = 5\)
c) không có
d) \(\sqrt {{5^2}} = 5\)
e) \(\sqrt {{{\left( { - 5} \right)}^2}} = \sqrt {25} = 5\)
e) \({2 \over 3}\sqrt {81} - \left( { - {3 \over 4}} \right):\sqrt {{9 \over {64}}} + {\left( {{{\sqrt 2 } \over 3}} \right)^0}\)\(\; - {\left( {\sqrt 3 } \right)^2}\)
\(\eqalign{ & = {2 \over 3}.9 - \left( { - {3 \over 4}} \right):{3 \over 8} + 1 - 3 \cr & = 6 - \left( { - {3 \over 4}} \right).{3 \over 8} - 2 \cr&= 6 + 2 - 2 = 6. \cr} \)
Bài 2:
a) \(\left| x \right| = \sqrt 2 \Rightarrow x = \pm \sqrt 2 \)
b \(\left| {x - \sqrt 2 } \right| = \sqrt 3 - 1 \)
\(\Rightarrow x - \sqrt 2 = \sqrt 3 - 1\) hoặc \(x - \sqrt 2 = - \left({\sqrt 3 - 1} \right)\)
\( \Rightarrow x = \sqrt 3 - 1 + \sqrt 2 \) hoặc \(x = \sqrt 2 - \sqrt 3 + 1.\)
Bài 3:
a) Ta có: \( - 3 = - \sqrt 9 > - \sqrt {10} \),
Vậy \( - 3 > - \sqrt {10} .\)
b) Ta có: \(\left. \matrix{ \sqrt {2009} > \sqrt {2008} \hfill \cr \sqrt {2006} < \sqrt {2007} \hfill \cr} \right\} \)\(\Rightarrow \sqrt {2009} - \sqrt {2006} > \sqrt {2008} - \sqrt {2007} \)
