Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 36

\[\boxed{\mathbf{36.}}\]

\[\textbf{а)}\lg(8x + 11) + \sqrt{x} =\]

\[= \lg{(x^{2} + 2)} + \sqrt{x}\]

\[x \geq 0;\]

\[8x + 11 > 0\]

\[x > - \frac{11}{8}.\]

\[M = \lbrack 0; + \infty).\]

\[\lg{(8x + 11)} = \lg\left( x^{2} + 2 \right)\]

\[8x + 11 = x^{2} + 2\]

\[x^{2} - 8x - 9 = 0\]

\[D_{1} = 16 + 9 = 25\]

\[x_{1} = 4 + 5 = 9;\]

\[x_{2} = 4 - 5 = - 1 < 0.\]

\[Ответ:x = 9.\]

\[\textbf{б)}\lg(x + 8) + \sqrt{- x} =\]

\[= \lg{(x^{2} + 2)} + \sqrt{- x}\]

\[- x \geq 0\]

\[x \leq 0.\]

\[x + 8 > 0\]

\[x > - 8.\]

\[x^{2} + 2 > 0\]

\[x^{2} > - 2\]

\[x \in R.\]

\[M = ( - 8;0\rbrack.\]

\[x + 8 = x^{2} + 2\]

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

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

\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 3 > 0;\ \ \]

\[x_{2} = - 2.\]

\[Ответ:x = - 2.\]