Решебник по алгебре 9 класс Мерзляк Задание 135

 

\[\boxed{\text{135\ (135).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 3(4x + 9) + 15 > 7(8 - x)\]

\[12x + 27 + 5 > 56 - 7x\]

\[19x > 56 - 32\]

\[19x > 24\]

\[x > \frac{24}{19}\]

\[Ответ:x \in \left( 1\frac{5}{19};\ + \infty \right).\]

\[2)\ (2 - y)(3 + y) \leq\]

\[\leq (4 + y)(6 - y)\ \]

\[6 + 2y - 3y - y^{2} \leq\]

\[\leq 24 - 4y + 6y - y^{2}\]

\[- y - y^{2} - 2y + y^{2} \leq 24 - 6\]

\[- 3y \leq 18\]

\[y \geq - 6\]

\[Ответ:x \in \lbrack - 6;\ + \infty).\]

\[3)\ (y + 3)(y - 5) -\]

\[- (y - 1)^{2} > - 16\]

\[y^{2} - 5y + 3y - 15 -\]

\[- y^{2} + 2y - 1 + 16 > 0\]

\[0 > 0 - неверно.\]

\[Ответ:\ \varnothing.\]

\[4)\ \frac{3x - 7}{5} - 1 \geq \frac{2x - 6}{3}\ \ | \cdot 15\]

\[3 \cdot (3x - 7) - 15 \geq 5 \cdot (2x - 6)\]

\[9x - 21 - 15 - 10x + 30 \geq 0\]

\[- x - 6 \geq 0\]

\[- x \geq 6\]

\[x \leq - 6\]

\[Ответ:x \in ( - \infty; - 6\rbrack.\]

\[5)\ \frac{2x}{3} - \frac{x - 1}{6} - \frac{x + 2}{2} < 0\ | \cdot 6\]

\[4x - x + 1 - 3x - 6 < 0\]

\[0x < 5\]

\[Ответ:x \in ( - \infty;\ + \infty).\]

\[6)\ \frac{y - 1}{2} - \frac{2y + 1}{8} - y < 2\ \ | \cdot 8\]

\[4y - 4 - 2y - 1 - 8y < 16\]

\[- 6y - 5 < 16\]

\[- 6y < 21\]

\[y > - \frac{21}{6}\]

\[y > - 3\frac{3}{6}\]

\[y > - 3,5\]

\[Ответ:y \in ( - 3,5; + \infty).\]

\[\boxed{\mathbf{135.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x^{4} - 5x^{2} + 4 = 0\]

\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 5t + 4 = 0 \\ \end{matrix} \right.\ \)

\(t_{1} + t_{2} = 5,\ \ t_{1} \cdot t_{2} = 4,\ \ \)

\[t_{1} = 4,\ \ t_{2} = 1\]

\[\left\{ \begin{matrix} (t - 4)(t - 1) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ x = \pm 2;\ x = \pm 1.\]

\[2)\ x^{4} - 5x^{2} + 6 = 0\]

\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 5t + 6 = 0 \\ \end{matrix} \right.\ \)

\(t_{1} + t_{2} = 5,\ \ t_{1} \cdot t_{2} = 6,\ \ \)

\[t_{1} = 3,\ \ t_{2} = 2\]

\[Ответ:x = \pm \sqrt{3};\ x = \pm \sqrt{2}.\ \]

\[3)\ x^{4} - 8x^{2} - 9 = 0\]

\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 8t - 9 = 0 \\ \end{matrix} \right.\ \)

\[t_{1} + t_{2} = 8,\ \ t_{1} \cdot t_{2} = - 9,\ \ \]

\[t_{1} = 9,\ \ t_{2} = - 1\]

\[Ответ:x = 3;\ x = - 3.\ \]

\[4)\ x^{4} + 14x^{2} - 32 = 0\]

\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² + 14t - 32 = 0 \\ \end{matrix} \right.\ \)

\(t_{1} + t_{2} = - 14,\ \ t_{1} \cdot t_{2} = - 32,\ \ \)

\[t_{1} = - 16,\ \ t_{2} = 2\]

\[\left\{ \begin{matrix} (t + 16)(t - 2) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm \sqrt{2}\text{.\ }\]

\[5)\ 4x^{4} - 9x^{2} + 2 = 0\]

\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4t² - 9t + 2 = 0 \\ \end{matrix} \right.\ \)

\[D = 81 - 32 = 49,\ \ \]

\[t_{1} = \frac{9 + 7}{8} = 2,\ \ \]

\[t_{2} = \frac{9 - 7}{8} = \frac{1}{4}\]

\[\left\{ \begin{matrix} 4 \cdot (t - 2)\left( t - \frac{1}{4} \right) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm \frac{1}{2};x = \pm \sqrt{2}\text{.\ }\]

\[6)\ {3x}^{4} + 8x^{2} - 3 = 0\]

\[\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3t² + 8t - 3 = 0 \\ \end{matrix} \right.\ \]

\[D = 64 + 36 = 100,\ \ \]

\[t_{1} = \frac{- 8 + 10}{6} = \frac{1}{3},\ \ \]

\[t_{2} = \frac{- 8 - 10}{6} = - 3\]

\[\left\{ \begin{matrix} 3 \cdot \left( t - \frac{1}{3} \right)(t + 3) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm \frac{1}{\sqrt{3}}\text{.\ }\]