ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 612

\[\boxed{\mathbf{612}.}\]

\[1)\ \sqrt[3]{x - 2} = 2;\]

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

\[x - 2 = 8;\]

\[x = 8 + 2 = 10;\]

\[Ответ:\ \ x = 10.\]

\[2)\ \sqrt[3]{2x + 7} = \sqrt[3]{3(x - 1)};\]

\[2x + 7 = 3(x - 1);\]

\[2x + 7 = 3x - 3;\]

\[2x - 3x = - 3 - 7;\]

\[- x = - 10;\]

\[x = 10;\]

\[Ответ:\ \ x = 10.\]

\[3)\ \sqrt[4]{25x^{2} - 144} = x;\]

\[25x^{2} - 144 = x^{4};\]

\[x^{4} - 25x^{2} + 144 = 0;\]

\[Пусть\ y = x^{2}:\]

\[y^{2} - 25y + 144 = 0;\]

\[D = 25^{2} - 4 \bullet 144 =\]

\[= 625 - 576 = 49\]

\[y_{1} = \frac{25 - 7}{2} = 9\ \ и\ \ \]

\[y_{2} = \frac{25 + 7}{2} = 16;\]

\[x_{1} = \pm \sqrt{9} = \pm 3\ \ и\ \ \]

\[x_{2} = \pm \sqrt{16} = \pm 4;\]

\[Выполним\ проверку:\]

\[\sqrt[4]{25 \bullet ( \pm 3)^{2} - 144} =\]

\[= \sqrt[4]{225 - 144} = \sqrt[4]{81} = 3;\]

\[\sqrt[4]{25 \bullet ( \pm 4)^{2} - 144} =\]

\[= \sqrt[4]{400 - 144} = \sqrt[4]{256} = 4;\]

\[Ответ:\ \ x_{1} = 3;\ \ x_{2} = 4.\]

\[4)\ x^{2} = \sqrt{19x^{2} - 34};\]

\[x^{4} = 19x^{2} - 34;\]

\[x^{4} - 19x^{2} + 34 = 0;\]

\[Пусть\ y = x^{2}:\]

\[y^{2} - 19y + 34 = 0;\]

\[D = 19^{2} - 4 \bullet 34 =\]

\[= 361 - 136 = 225\]

\[y_{1} = \frac{19 - 15}{2} = 2\ \ и\]

\[\text{\ \ }y_{2} = \frac{19 + 15}{2} = 17;\]

\[x_{1} = \pm \sqrt{2}\text{\ \ }и\ \ x_{2} = \pm \sqrt{17};\]

\[Выполним\ проверку:\]

\[\sqrt{19 \bullet \left( \pm \sqrt{2} \right)^{2} - 34} =\]

\[= \sqrt{38 - 34} = \sqrt{4} =\]

\[= 2 = \left( \pm \sqrt{2} \right)^{2};\]

\[\sqrt{19 \bullet \left( \pm \sqrt{17} \right)^{2} - 34} =\]

\[= \sqrt{323 - 34} = \sqrt{289} =\]

\[= 17 = \left( \pm \sqrt{17} \right)^{2};\]

\[Ответ:\ \ x_{1} = \pm \sqrt{2};\ \ x_{2} = \pm \sqrt{17}.\]