ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 615

\[\boxed{\mathbf{615}.}\]

\[1)\ \sqrt{4x + 2\sqrt{3x^{2} + 4}} = x + 2;\]

\[4x + 2\sqrt{3x^{2} + 4} = x^{2} + 4x + 4;\]

\[2\sqrt{3x^{2} + 4} = x^{2} + 4;\]

\[4\left( 3x^{2} + 4 \right) = x^{4} + 8x^{2} + 16;\]

\[12x^{2} + 16 = x^{4} + 8x^{2} + 16;\]

\[x^{4} - 4x^{2} = 0;\]

\[x^{2} \bullet \left( x^{2} - 4 \right) = 0;\]

\[(x + 2) \bullet x^{2} \bullet (x - 2) = 0;\]

\[x_{1} = - 2,\ \ \ x_{2} = 0,\ \ \ x_{3} = 2;\]

\[Выполним\ проверку:\]

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

\[= \sqrt{- 8 + 2\sqrt{16}} + 2 =\]

\[= \sqrt{0} + 2 = 2;\]

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

\[= \sqrt{8 + 2\sqrt{16}} - 2 =\]

\[= \sqrt{16} - 2 = 2;\]

\[\sqrt{4 \bullet 0 + 2\sqrt{3 \bullet 0^{2} + 4}} - 0 =\]

\[= \sqrt{2\sqrt{4}} = \sqrt{4} = 2;\]

\[Ответ:\ \ x_{1} = \pm 2;\ \ x_{2} = 0.\]

\[2)\ 3 - x = \sqrt{9 - \sqrt{36x^{2} - 5x^{4}}};\]

\[9 - 6x + x^{2} =\]

\[= 9 - \sqrt{36x^{2} - 5x^{4}};\]

\[x^{2} - 6x = - \sqrt{36x^{2} - 5x^{4}};\]

\[x^{4} - 12x^{3} + 36x^{2} =\]

\[= 36x^{2} - 5x^{4};\]

\[6x^{4} - 12x^{3} = 0;\]

\[x^{4} - 2x^{3} = 0;\]

\[x^{3} \bullet (x - 2) = 0;\]

\[x_{1} = 0\ \ и\ \ x_{2} = 2;\]

\[Выполним\ проверку:\]

\[\sqrt{9 - \sqrt{36 \bullet 0^{2} - 5 \bullet 0^{4}}} + 0 =\]

\[= \sqrt{9 - \sqrt{0}} = 3;\]

\[\sqrt{9 - \sqrt{36 \bullet 2^{2} - 5 \bullet 2^{4}}} + 2 =\]

\[= \sqrt{9 - \sqrt{64}} + 2 = \sqrt{1} + 2 = 3;\]

\[Ответ:\ \ x_{1} = 0;\ \ x_{2} = 2.\]

\[3)\ \sqrt{x^{2} + 3x + 12} -\]

\[- \sqrt{x^{2} + 3x} = 2;\]

\[\sqrt{x^{2} + 3x + 12} =\]

\[= 2 + \sqrt{x^{2} + 3x};\]

\[x^{2} + 3x + 12 =\]

\[= 4 + 4\sqrt{x^{2} + 3x} + x^{2} + 3x;\]

\[8 = 4\sqrt{x^{2} + 3x};\]

\[2 = \sqrt{x^{2} + 3x};\]

\[4 = x^{2} + 3x;\]

\[x^{2} + 3x - 4 = 0;\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 =\]

\[= 25,\ тогда:\]

\[x_{1} = \frac{- 3 - 5}{2} = - 4\ \ и\ \ \]

\[x_{2} = \frac{- 3 + 4}{2} = 1;\]

\[Выполним\ проверку:\]

\[\sqrt{16 - 3 \bullet 4 + 12} - \sqrt{16 - 3 \bullet 4} =\]

\[= \sqrt{16} - \sqrt{4} = 4 - 2 = 2;\]

\[\sqrt{1 + 3 + 12} - \sqrt{1 + 3} =\]

\[= \sqrt{16} - \sqrt{4} = 4 - 2 = 2;\]

\[Ответ:\ \ x_{1} = - 4;\ \ x_{2} = 1.\]

\[4)\ \sqrt{x^{2} + 5x + 10} -\]

\[- \sqrt{x^{2} + 5x + 3} = 1;\]

\[\sqrt{x^{2} + 5x + 10} =\]

\[= \sqrt{x^{2} + 5x + 3} + 1;\]

\[x^{2} + 5x + 10 = x^{2} + 5x +\]

\[+ 3 + 2\sqrt{x^{2} + 5x + 3} + 1;\]

\[6 = 2\sqrt{x^{2} + 5x + 3};\]

\[3 = \sqrt{x^{2} + 5x + 3};\]

\[9 = x^{2} + 5x + 3;\]

\[x^{2} + 5x - 6 = 0;\]

\[D = 5^{2} + 4 \bullet 6 = 25 + 24 = 49\]

\[x_{1} = \frac{- 5 - 7}{2} = - 6\ \ и\]

\[\text{\ \ }x_{2} = \frac{- 5 + 7}{2} = 1;\]

\[Выполним\ проверку:\]

\[\sqrt{36 - 5 \bullet 6 + 10} -\]

\[- \sqrt{36 - 5 \bullet 6 + 3} = \sqrt{16} - \sqrt{9} =\]

\[= 4 - 3 = 1;\]

\[\sqrt{1 + 5 + 10} - \sqrt{1 + 5 + 3} =\]

\[= \sqrt{16} - \sqrt{9} = 4 - 3 = 1;\]

\[Ответ:\ \ x_{1} = - 6;\ \ x_{2} = 1.\]