Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1595

\[\boxed{\mathbf{1595}\mathbf{.}}\]

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

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

\[= 3x - 2\]

\[x + 3 + 2x - 4 - 3x + 2 =\]

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

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

\[1 = 4(x + 3)(2x - 4)\]

\[4\left( 2x^{2} - 4x + 6x - 12 \right) - 1 = 0\]

\[8x^{2} + 8x - 48 - 1 = 0\]

\[8x^{2} + 8x - 49 = 0\]

\[D = 64 + 1568 = 1632 = 16 \bullet 102\]

\[x = \frac{- 8 \pm \sqrt{1632}}{2 \bullet 8} =\]

\[= \frac{- 8 \pm 4\sqrt{102}}{16} = \frac{- 2 \pm \sqrt{102}}{4}.\]

\[Имеет\ смысл\ при:\]

\[x + 3 \geq 0 \rightarrow x \geq - 3;\]

\[2x - 4 \geq 0 \rightarrow x \geq 2;\]

\[3x - 2 \geq 0 \rightarrow x \geq \frac{2}{3}.\]

\[Под\ знаком\ корня:\]

\[100 < 102 < 121\]

\[10 < \sqrt{102} < 11.\]

\[Ответ:\ \ x = \frac{\sqrt{102} - 2}{4}.\]

\[2)\ \frac{1}{1 - \sqrt{1 - x}} + \frac{1}{1 + \sqrt{1 - x}} =\]

\[= \frac{2\sqrt{2}}{\sqrt{1 - x}}\]

\[y = \sqrt{1 - x}:\]

\[\frac{1}{1 - y} + \frac{1}{1 + y} = \frac{2\sqrt{2}}{y}\ \ \ \ \ | \bullet y(1 - y)(1 + y)\]

\[y(1 + y) + y(1 - y) =\]

\[= 2\sqrt{2}(1 - y)(1 + y)\]

\[y + y^{2} + y - y^{2} = 2\sqrt{2}\left( 1 - y^{2} \right)\]

\[2y = 2\sqrt{2} - 2\sqrt{2}y^{2}\]

\[2\sqrt{2}y^{2} + 2y - 2\sqrt{2} = 0\ \ \ \ \ |\ :\sqrt{2}\]

\[2y^{2} + \sqrt{2}y - 2 = 0\]

\[D = 2 + 16 = 18 = 9 \bullet 2\]

\[y_{1} = \frac{- \sqrt{2} - \sqrt{18}}{2 \bullet 2} =\]

\[= \frac{- \sqrt{2} - 3\sqrt{2}}{4} = - \frac{4\sqrt{2}}{4} = - \sqrt{2};\]

\[y_{2} = \frac{- \sqrt{2} + \sqrt{18}}{2 \bullet 2} =\]

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

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

\[корней\ нет.\]

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

\[1 - x = \frac{2}{4}\]

\[x = 1 - \frac{2}{4} = \frac{1}{2}.\]

\[Имеет\ смысл\ при:\]

\[1 - x > 0\]

\[x < 1.\]

\[Ответ:\ \ x = \frac{1}{2}.\]