Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 43

\[\boxed{\mathbf{43.}}\]

\[\textbf{а)}\ \left( x^{2} + 1 \right)^{\sqrt{8x - x^{2} - 15}} =\]

\[= (2x + 9)^{\sqrt{8x - x^{2} - 15}}\]

\[2x + 9 > 0\]

\[2x > - 9\]

\[x > - 4,5.\]

\[8x - x^{2} - 15 \geq 0\]

\[x^{2} - 8x + 15 \leq 0\]

\[D_{1} = 16 - 15 = 1\]

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

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

\[(x - 3)(x - 5) \leq 0\]

\[3 \leq x \leq 5.\]

\[M = \lbrack 3;5\rbrack.\]

\[1)\ \sqrt{8x - x^{2} - 15} = 0\]

\[8x - x^{2} - 15 = 0\]

\[x^{2} - 8x + 15 = 0\]

\[D_{1} = 16 - 15 = 1\]

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

\[x_{2} = 4 - 1 = 3.\]

\[2)\lg{(x^{2} + 1)} = \lg{(2x + 9)}\]

\[x^{2} + 1 = 2x + 9\]

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

\[D_{1} = 1 + 8 = 9\]

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

\[x_{2} = 1 - 3 = - 2 < 3.\]

\[Ответ:x = 3;x = 4;x = 5.\]

\[\textbf{б)}\ \left( x^{2} - 8 \right)^{\sqrt{24x - 4x^{2} - 35}} =\]

\[= (7x - 20)^{\sqrt{24x - 4x^{2} - 35}}\]

\[x^{2} - 8 > 0\]

\[\left( x + \sqrt{8} \right)\left( x - \sqrt{8} \right) > 0\]

\[x < - \sqrt{8};\ \ x > \sqrt{8}.\]

\[7x - 20 > 0\]

\[7x > 20\]

\[x > \frac{20}{7} > 2\frac{6}{7}.\]

\[24x - 4x^{2} - 35 \geq 0\]

\[4x^{2} - 24x + 35 \leq 0\]

\[D_{1} = 144 - 140 = 4\]

\[x_{1} = \frac{12 + 2}{4} = \frac{14}{4} = \frac{7}{2} = 3,5;\]

\[x_{2} = \frac{12 - 2}{4} = \frac{10}{4} = \frac{5}{2} = 2,5.\]

\[\ (x - 2,5)(x - 3,5) \leq 0\]

\[2,5 \leq x \leq 3,5.\]

\[M = \left( 2\frac{6}{7};3,5 \right\rbrack.\]

\[1)\ \sqrt{24x - 4x^{2} - 35} = 0\]

\[- 4x^{2} + 24x - 35 = 0\]

\[4x^{2} - 24x + 35 = 0\]

\[D_{1} = 144 - 140 = 4\]

\[x_{1} = \frac{12 + 2}{4} = \frac{14}{4} = \frac{7}{2} = 3,5;\]

\[x_{2} = \frac{12 - 2}{4} = \frac{10}{4} = \frac{5}{2} =\]

\[= 2,5 < 2\frac{6}{7}.\]

\[2)\lg\left( x^{2} - 8 \right) = \lg(7x - 20)\]

\[x^{2} - 8 = 7x - 20\]

\[x^{2} - 7x + 12 = 0\]

\[x_{1} + x_{2} = 7;\ \ x_{1} \cdot x_{2} = 12\]

\[x_{1} = 3;\ \ x_{2} = 4 > 3,5.\]

\[Ответ:x = 3,5;x = 3.\]