Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 11

\[\boxed{\mathbf{11.}}\]

\[\textbf{а)}\ \sqrt{x^{3} - 5x^{2} + 7x - 17} =\]

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

\[- x^{2} + 10x - 21 = 0\]

\[x^{2} - 10x + 21 = 0\]

\[D_{1} = 25 - 21 = 4\]

\[x_{1} = 5 + 2 = 7;\]

\[x_{2} = 5 - 2 = 3.\]

\[2)\ Подставим\ в\ неравенство.\]

\[x = 3:\]

\[3^{3} - 5 \cdot 3^{2} + 7 \cdot 3 - 17 =\]

\[= 27 - 45 + 21 - 17 = - 14 < 0\]

\[x = 3 - не\ корень.\]

\[x = 7:\]

\[7^{3} - 5 \cdot 7^{2} + 7 \cdot 7 - 17 =\]

\[= 343 - 245 + 49 - 17 =\]

\[= 130 > 0\]

\[x = 7 - корень.\]

\[Ответ:x = 7.\]

\[\textbf{б)}\ \sqrt{x^{3} - 8x^{2} - 7x + 2} =\]

\[= \sqrt{x^{3} - 7x^{2} - 18x + 20}\]

\[- x^{2} + 11x - 18 = 0\]

\[x^{2} - 11x + 18 = 0\]

\[x_{1} + x_{2} = 11;\ \ \ x_{1} \cdot x = 18\]

\[x_{1} = 2;\ \ x_{2} = 9.\]

\[2)\ Подставим\ в\ неравенство.\]

\[x = 2:\]

\[2^{3} - 8 \cdot 2^{2} - 7 \cdot 2 + 2 =\]

\[= 8 - 32 - 14 + 2 =\]

\[= - 36 < 0;\]

\[x = 2 - не\ корень.\]

\[x = 9:\]

\[9^{3} - 8 \cdot 9^{2} - 7 \cdot 9 + 9 = 2 =\]

\[= 729 - 648 - 63 + 2 = 20 > 0;\]

\[x = 9 - корень.\]

\[Ответ:x = 9.\]