Cho \(A = \) \(\left[ { - \sqrt {2,25}  + 4\sqrt {{{\left( { - 2,15} \right)}^2}}  - {{\left( {3\sqrt {\dfrac{7}{6}} } \right)}^2}} \right] .\sqrt {1\dfrac{9}{{16}}}\) và $B = 1,68 + \left[ {\dfrac{4}{5} - 1,2\left( {\dfrac{5}{2} - 1\dfrac{3}{4}} \right)} \right]:\left[ {{{\left( {\dfrac{2}{3}} \right)}^2} + \dfrac{1}{9}} \right].$ So sánh \(A\) và \(B\).

Ta có

\(A = \left[ { - \sqrt {2,25}  + 4\sqrt {{{\left( { - 2,15} \right)}^2}}  - {{\left( {3\sqrt {\dfrac{7}{6}} } \right)}^2}} \right].\sqrt {1\dfrac{9}{{16}}} \)

\(A = \left[ { - 1,5 + 4.2,15 - 9.\dfrac{7}{6}} \right].\sqrt {\dfrac{{25}}{{16}}} \)

\(A = \left[ { - 1,5 + 8,6 - \dfrac{{21}}{2}} \right].\dfrac{5}{4}\)

\(A = \left[ {7,1 - 10,5} \right].1,25\)

\(A =  - 3,4.1,25\)

\(A =  - 4,25\)

Và

$B = 1,68 + \left[ {\dfrac{4}{5} - 1,2\left( {\dfrac{5}{2} - 1\dfrac{3}{4}} \right)} \right]:\left[ {{{\left( {\dfrac{2}{3}} \right)}^2} + \dfrac{1}{9}} \right]$

$B = \dfrac{{42}}{{25}} + \left[ {\dfrac{4}{5} - \dfrac{6}{5}\left( {\dfrac{5}{2} - \dfrac{7}{4}} \right)} \right]:\left[ {\dfrac{4}{9} + \dfrac{1}{9}} \right]$

$B = \dfrac{{42}}{{25}} + \left[ {\dfrac{4}{5} - \dfrac{6}{5}.\dfrac{3}{4}} \right]:\dfrac{5}{9}$

$B = \dfrac{{42}}{{25}} + \left[ {\dfrac{4}{5} - \dfrac{9}{{10}}} \right]:\dfrac{5}{9}$

$B = \dfrac{{42}}{{25}} + \dfrac{{ - 1}}{{10}}:\dfrac{5}{9} = \dfrac{{42}}{{25}} + \dfrac{{ - 9}}{{50}}$

$B = \dfrac{{84}}{{50}} + \dfrac{{ - 9}}{{50}} = \dfrac{{75}}{{50}} = \dfrac{3}{2}$

Từ đó \(A < B\).