ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 463

\[\boxed{\mathbf{463}.}\]

\[1)\ \sqrt{4 + 2\sqrt{3}} - \sqrt{4 - 2\sqrt{3}} = 2\]

\[Возведем\ обе\ части\ равенства\]

\[\ во\ вторую\ степень:\]

\[\left( 4 + 2\sqrt{3} \right) - 2\sqrt{4 + 2\sqrt{3}} \bullet\]

\[\bullet \sqrt{4 - 2\sqrt{3}} + \left( 4 - 2\sqrt{3} \right) = 4\]

\[4 + 2\sqrt{3} + 4 - 2\sqrt{3} -\]

\[- 2\sqrt{\left( 4 + 2\sqrt{3} \right)\left( 4 - 2\sqrt{3} \right)} = 4\]

\[8 - 2\sqrt{16 - 4 \bullet 3} = 4\]

\[8 - 2\sqrt{16 - 12} = 4\]

\[8 - 2\sqrt{4} = 4\]

\[8 - 2 \bullet 2 = 4\]

\[8 - 4 = 4\]

\[Тождество\ доказано.\]

\[2)\ \sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} = 3\]

\[Пусть\ x = \sqrt[3]{9 + \sqrt{80}} +\]

\[+ \sqrt[3]{9 - \sqrt{80}}:\]

\[x^{3} = \left( 9 + \sqrt{80} \right) +\]

\[+ 3\sqrt[3]{\left( 9 + \sqrt{80} \right)^{2}\left( 9 - \sqrt{80} \right)} +\]

\[+ 3\sqrt[3]{\left( 9 + \sqrt{80} \right)\left( 9 - \sqrt{80} \right)^{2}} +\]

\[+ \left( 9 - \sqrt{80} \right)\]

\[x^{3} = 18 + 3\sqrt[3]{81 - 80} \bullet\]

\[\bullet \left( \sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} \right)\]

\[x^{3} = 18 + 3 \bullet 1 \bullet x\]

\[x^{3} - 3x - 18 = 0\]

\[\left( x^{3} + 3x^{2} + 6x \right) -\]

\[- \left( 3x^{2} + 9x + 18 \right) = 0\]

\[x\left( x^{2} + 3x + 6 \right) -\]

\[- 3\left( x^{2} + 3x + 6 \right) = 0\]

\[(x - 3)\left( x^{2} + 3x + 6 \right) = 0\]

\[x - 3 = 0\]

\[\ x = 3.\]

\[Тождество\ доказано.\]