Question 39: Let \(\begin{array}{l} A = \left[ { – \sqrt {2,25} + 4\sqrt {{{\left( { – 2,15} \right)}^2}} – {{\left( {3\sqrt {\frac{7}{6}} } \right)}^2}} \right].\sqrt {1\frac{9}{{16}}} \\ B = 1.68 + \left[ {\frac{4}{5} – 1,2\left( {\frac{5}{2} – 1\frac{3}{4}} \right)} \right]:\left[ {{{\left( {\frac{2}{3}} \right)}^2} + \frac{1}{9}} \right]
\end{array}\). Compare A and B.
We have
\(\begin{array}{l} A = \left[ { – \sqrt {2,25} + 4\sqrt {{{\left( { – 2,15} \right)}^2}} – {{\left( {3\sqrt {\frac{7}{6}} } \right)}^2}} \right].\sqrt {1\frac{9}{{16}}} = \left[ { – 1,5 + 4.2,15 – 9.\frac{7}{6}} \right].\sqrt {\frac{{25}}{{16}}} A = \left[ { – 1,5 + 8,6 – \frac{{21}}{2}} \right].\frac{5}{4}A = \left[ {7,1 – 10,5} \right].1.25A = – 3,4.1,25A = – 4.25\\ B = 1.68 + \left[ {\frac{4}{5} – 1,2\left( {\frac{5}{2} – 1\frac{3}{4}} \right)} \right]:\left[ {{{\left( {\frac{2}{3}} \right)}^2} + \frac{1}{9}} \right] = \frac{{42}}{{25}} + \left[ {\frac{4}{5} – \frac{6}{5}\left( {\frac{5}{2} – \frac{7}{4}} \right)} \right]:\left[ {\frac{4}{9} + \frac{1}{9}} \right] = \frac{{42}}{{25}} + \left[ {\frac{4}{5} – \frac{6}{5}.\frac{3}{4}} \right]:\frac{5}{9} = B = \frac{{42}}{{25}} + \frac{{ – 1}}{{10}}:\frac{5}{9} = \frac {{42}}{{25}} + \frac{{ – 9}}{{50}} = \frac{{84}}{{50}} + \frac{{ – 9}}{{50} } = \frac{{75}}{{50}} = \frac{3}{2} \end{array}\)
Since then A
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