Bài 1:
a) \(\sqrt {0,49} = 0,7\)
b) \( - \sqrt {1,44} = - 1,2\)
c) \(\sqrt {{{10}^4}} = {10^2} = 100\)
d) \(\sqrt {{{0,09} \over {121}}} = {{0,3} \over {11}}\)
e) \({\left( { - \sqrt {{5 \over 4}} } \right)^2} - \sqrt {{9 \over 4}} :\left( { - 4,5} \right)\)\(\; - \sqrt {{{25} \over {16}}} .\sqrt {{{64} \over 9}} \)
\(\eqalign{ & = {5 \over 4} - {3 \over 2}:\left( {{{ - 45} \over {10}}} \right) - {5 \over 4}.{8 \over 3} \cr&= {5 \over 4} - {3 \over 2}\left( {{{ - 2} \over 9}} \right) - {{10} \over 3} \cr & = {5 \over 4} + {1 \over 3} - {{10} \over 3} = - {7 \over 4} \cr} \)
Bài 2:
a) \({x^2} = 9 \Rightarrow x = \pm 3.\)
b) \({x^2} - {{16} \over {25}} = 0 \Rightarrow {x^2} = {{16} \over {25}} \Rightarrow x = \pm {4 \over 5}.\)
c) \({x^2} + 1 = 0 \Rightarrow {x^2} = - 1\) ( không có x).
d) \({x^2} - 3 = 0 \Rightarrow {x^2} = 3 \Rightarrow x = \pm \sqrt 3 .\)
Bài 3:
a) \(6 = \sqrt {36} > \sqrt {35} \) vậy \(6 > \sqrt {35} \)
b) \(\sqrt 2 < \sqrt 3 \)
\(\sqrt {11} < \sqrt {25} = 5\).
Vậy \(\sqrt 2 + \sqrt {11} < \sqrt 3 + 5.\)
