Bài 1:
a) \({1 \over 2}.\sqrt {169} - \sqrt {{1 \over {16}}} \)
\(\;= {1 \over 2}.13 - {1 \over 4} = {{13} \over 2} - {1 \over 4} = {{25} \over 4}.\)
b) \(\left[ {0,{{\left( 5 \right)}^2}} \right] - \sqrt 2 + {3 \over 2} \)
\(\;= {\left( {{5 \over 9}} \right)^2} - \sqrt 2 + {3 \over 2}\)
\(\;= {{25} \over {81}} + {3 \over 2} - \sqrt 2 = {{293} \over {162}} - \sqrt 2 .\)
Bài 2:
a) Ta có: \(0,\left( {21} \right) = {{21} \over {99}};\,0,21 = {{21} \over {100}}.\)
Vì \(99 < 100 \Rightarrow {{21} \over {99}} > {{21} \over {100.}}\)
b) Ta có: \(1,\left( {4142} \right) = 1,41424142...\)
\(\sqrt 2 = 1,414213562...\)
\( \Rightarrow 1,\left( {4142} \right) > \sqrt 2 .\)
Bàu 3:
a) Ta có: \(1,\left( 4 \right) = {{13} \over 9}.\)
Vậy \(\left| {x - \sqrt 2 } \right| = {{13} \over 9}\)
\(\Rightarrow x - \sqrt 2 = {{13} \over 9}\) hoặc \(x - \sqrt 2 = - {{13} \over 9}\)
\(\Rightarrow x = \sqrt 2 + {{13} \over 9}\) hoặc \(x = \sqrt 2 - {{13} \over 9}\)
b) Ta có : \(0,\left( {71} \right) = {{71} \over {99}}\)
Vậy : \(\left| {1 - x} \right| = \sqrt 3 - {{71} \over {99}} \)
\(\Rightarrow 1 - x = \sqrt 3 - {{71} \over {99}}\) hoặc \(1 - x = - \sqrt 3 + {{71} \over {99}}\)
\(\Rightarrow x = {{170} \over {99}} - \sqrt 3 \) hoặc \(x = {{28} \over {99}} + \sqrt 3 \).
