Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 15

\[\boxed{\mathbf{15}\mathbf{.}}\]

\[\textbf{а)}\ \frac{2\sqrt{0,5 - x}}{(x - 4)(x + 1)} =\]

\[= \frac{\sqrt{0,5 - x}}{(x - 2)(x + 2)}\]

\[1)\ x \neq 4;x \neq - 1;\]

\[x \neq 2;\ \ x \neq - 2.\]

\[0,5 - x \geq 0\]

\[x \leq 0,5.\]

\[\sqrt{0,5 - x}\left( x^{2} + 3x - 4 \right) = 0\]

\[\sqrt{0,5 - x} = 0\]

\[0,5 - x = 0\]

\[x = 0,5.\]

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

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

\[x_{1} = - 4;\ \ \]

\[x_{2} = 1\ (не\ подходит).\]

\[Ответ:x = - 4;x = 0,5.\]

\[\textbf{б)}\ \frac{4\sqrt{x - 10}}{(x - 4)(x + 3)} =\]

\[= \frac{3\sqrt{x - 10}}{(x - 3)(x - 2)}\]

\[1)\ x \neq 4;x \neq - 3;\]

\[x \neq 3;x \neq 2.\]

\[x - 10 \geq 0\]

\[x \geq 10.\]

\[\sqrt{x - 10}\left( x^{2} - 17x + 60 \right) = 0\]

\[\sqrt{x - 10} = 0\]

\[x - 10 = 0\]

\[x = 10.\]

\[x^{2} - 17x + 60 = 0\]

\[x_{1} + x_{2} = 17;\ \ x_{1} \cdot x_{2} = 60\]

\[x_{1} = 5\ (не\ подходит);\]

\[x_{2} = 12.\]

\[Ответ:x = 10;x = 12.\]