ГДЗ по алгебре 9 класс Рурукин контрольные работы КР-3. Уравнения и неравенства с одной переменной Вариант 3

Условие:

1. Решите уравнение (x^2+27x-57)^2=(x^2-3x+1)^2.

2. Найдите решения уравнения (3x+2)/(2x+3)+(2x+3)/(3x+2)+2=0.

3. Найдите корни уравнения (x^2-16) √(x+2)=0.

4. Решите неравенство (x-2)(3x^2-5x-2)(x+4)≥0.

5. Определите значения a, при которых уравнение x³+6x²+ax=0 имеет два корня. Найдите эти корни.

6. Решите неравенство (x^2-3)(2x^2-3x+1)<(x^2-7)(2x^2-3x+1).

\[\boxed{\mathbf{1}\mathbf{.}\mathbf{\ }}\]

\[\left( x^{2} + 27x - 57 \right)^{2} = \left( x^{2} - 3x + 1 \right)^{2}\]

\[1)\ x^{2} + 27x - 57 = x^{2} - 3x + 1\]

\[27x + 3x = 1 + 57\]

\[30x - 58 = 0\]

\[2 \cdot (15x - 29) = 0\]

\[15x = 29\]

\[x = \frac{29}{15}.\]

\[2)\ x^{2} + 27x - 57 = - x^{2} + 3x - 1\]

\[2x^{2} + 24x - 56 = 0\]

\[2 \cdot \left( x^{2} + 15x - 28 \right) = 0\]

\[2 \cdot (x - 2)(x + 14) = 0\]

\[x = 2;\ \ \ x = - 14\]

\[Ответ:x = \frac{29}{15};\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\ x = 2;\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }x = - 14.\]

\[\boxed{\mathbf{2}\mathbf{.}\mathbf{\ }}\]

\[ОДЗ:\ \ x \neq - 1,5;\ \ x \neq - \frac{2}{3}\]

\[13x^{2} + 24x + 13 + 12x^{2} + 26x + 12 = 0\]

\[25x^{2} + 50x + 25 = 0\ \ \ \ |\ :25\]

\[x^{2} + 2x + 1 = 0\]

\[(x + 1)^{2} = 0\]

\[x + 1 = 0\]

\[x = - 1.\]

\[Ответ:x = - 1.\]

\[\boxed{\mathbf{3}\mathbf{.}\mathbf{\ }}\]

\[\left( x^{2} - 16 \right)\sqrt{x + 2} = 0\]

\[x^{2} - 16 = 0\]

\[(x - 4)(x + 4) = 0\]

\[x = 4;\ \ x = - 4.\]

\[\sqrt{x + 2} \geq 0\]

\[x + 2 \geq 0\]

\[x \geq - 2.\]

\[Ответ:x = 4;\ \ x = - 2.\]

\[\boxed{\mathbf{4}\mathbf{.}\mathbf{\ }}\]

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

\[3x^{2} - 5x - 2 = 3 \cdot \left( x + \frac{1}{3} \right)(x - 2)\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 + 7}{6} = 2;\ \ \]

\[x_{2} = \frac{5 - 7}{6} = - \frac{1}{3}\]

\[3(x + 4)\left( x + \frac{1}{3} \right)(x - 2)(x - 2) \geq 0\]

\[Ответ:\]

\[x \in ( - \infty; - 4\rbrack \cup \left\lbrack - \frac{1}{3}; + \infty \right).\]

\[\boxed{\mathbf{5}\mathbf{.}\mathbf{\ }}\]

\[x^{3} + 6x^{2} + ax = 0\]

\[x\left( x^{2} + 6x + a \right) = 0\]

\[x^{2} + 6x + a = 0\]

\[D = 36 - 4a\]

\[36 - 4a = 0\]

\[- 4a = - 36\]

\[a = 9.\]

\[при\ a = 0:\]

\[x^{3} + 6x^{2} = 0\]

\[x^{2}(x + 6) = 0\]

\[x = 0;\ \ x = - 6.\]

\[при\ a = 9:\]

\[x^{3} + 6x^{2} + 9x = 0\]

\[x\left( x^{2} + 6x + 9 \right) = 0\]

\[x(x + 3)^{2} = 0\]

\[x = 0;\ \ x = - 3.\]

\[\boxed{\mathbf{6}\mathbf{.}\mathbf{\ }}\]

\[8x^{2} - 12x + 4 < 0\]

\[4 \cdot \left( 2x^{2} - 3x + 1 \right) < 0\]

\[2x^{2} - 3x + 1 =\]

\[D = 9 - 8 = 1\]

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

\[x_{2} = \frac{3 - 1}{4} = 0,5\]

\[4 \cdot (x - 0,5)(x - 4) < 0\]

\[Ответ:x \in (0,5;1).\]