Задача 94. Вычислите \(\sqrt {\,\left| {\,24\sqrt 3 -43\,} \right|} -\sqrt {24\sqrt 3 + 43} \)
ОТВЕТ: -8.
\(\sqrt {\left| {24\sqrt 3 -43} \right|} -\sqrt {24\sqrt 3 + 43} .\) Так как \(24\sqrt 3 = \sqrt {1728} < \sqrt {1849} = 43\), то \(\left| {24\sqrt 3 -43} \right| = 43-24\sqrt 3 .\) \(\sqrt {\left| {24\sqrt 3 -43} \right|} -\sqrt {24\sqrt 3 + 43} = \sqrt {43-24\sqrt 3 } -\sqrt {24\sqrt 3 + 43} = \) \( = \sqrt {27-24\sqrt 3 + 16} -\sqrt {27 + 24\sqrt 3 + 16} = \) \( = \sqrt {{{\left( {3\sqrt 3 } \right)}^2}-2 \cdot 3\sqrt 3 \cdot 4 + {4^2}} -\sqrt {{{\left( {3\sqrt 3 } \right)}^2} + 2 \cdot 3\sqrt 3 \cdot 4 + {4^2}} = \) \( = \sqrt {{{\left( {3\sqrt 3 -4} \right)}^2}} -\sqrt {{{\left( {3\sqrt 3 + 4} \right)}^2}} = \left| {3\sqrt 3 -4} \right|-\left| {3\sqrt 3 + 4} \right|.\) Так как \(3\sqrt 3 = \sqrt {27} > \sqrt {16} = 4\), то \(3\sqrt 3 -4 > 0\) и \(\left| {3\sqrt 3 -4} \right| = 3\sqrt 3 -4.\) \(\left| {3\sqrt 3 -4} \right|-\left| {3\sqrt 3 + 4} \right| = 3\sqrt 3 -4-3\sqrt 3 -4 = -8.\) Ответ: \(-8.\)