ГДЗ по алгебре 9 класс Рурукин контрольные работы КР-3. Уравнения и неравенства с одной переменной Вариант 5

Условие:

1. Решите уравнение x^4-x^3-7x^2+x+6=0.

2. Найдите решения уравнения (x-2x/(x+2))^2=5-(4x^2)/(x+2).

3. Найдите корни уравнения (x-2)(x-1)(x+2)(x+3)=60.

4. Решите неравенство x^2-3|x+1|+2x≤-1.

5. Докажите, что уравнение (x²+2x+2)(x²-4x+5)=1 не имеет корней.

6. Решите неравенство 3/(x^2+8x+17)+4/(x^2+8x+18)≥5.

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

\[x^{4} - x^{3} - 7x^{2} + x + 6 = 0\]

\[x = 1.\]

\[Разделим\ на\ (x - 1):\]

\[x^{3} - 7x - 6 = 0\]

\[x = - 1.\]

\[Разделим\ на\ (x + 1):\]

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

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

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

\[Ответ:x = \pm 1;x = - 2;x = 3.\]

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

\[\left( x^{\backslash x + 2} - \frac{2x}{x + 2} \right)^{2} = 5 - \frac{4x^{2}}{x + 2}\]

\[\left( \frac{x^{2} + 2x - 2x}{x + 2} \right)^{2} = 5 - 4 \cdot \frac{x²}{x + 2}\]

\[\left( \frac{x^{2}}{x + 2} \right)^{2} = 5 - 4 \cdot \frac{x²}{x + 2}\]

\[y = \frac{x^{2}}{x + 2}:\]

\[y^{2} = 5 - 4y\]

\[y^{2} + 4y - 5 = 0\]

\[D = 4 + 5 = 9\]

\[y_{1} = - 2 + 3 = 1;\ \ \]

\[y_{2} = - 2 - 3 = - 5\]

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

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

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

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

\[2)\frac{x^{2}}{x + 2} = - 5\]

\[x^{2} + 5x + 10 = 0\]

\[D = 25 - 40 < 0\]

\[нет\ корней.\]

\[Ответ:x = - 1;x = 2.\]

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

\[(x - 2)(x - 1)(x + 2)(x + 3) = 60\]

\[\left( (x - 2)(x + 3) \right)\left( (x - 1)(x + 2) \right) = 60\]

\[\left( x^{2} + x - 6 \right)\left( x^{2} + x - 2 \right) = 60\]

\[y = x^{2} + x - 6:\]

\[y(y + 4) = 60\]

\[y^{2} + 4y - 60 = 0\]

\[D = 4 + 60 = 64\]

\[y_{1} = - 2 + 8 = 6;\ \ \]

\[y_{2} = - 2 - 8 = - 10\]

\[1)\ y = - 10:\]

\[x^{2} + x - 6 = - 10\]

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

\[D = 1 - 16 < 0\]

\[нет\ корней.\]

\[2)\ y = 6:\]

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

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

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

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

\[Ответ:x = - 4;\ \ x = 3.\]

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

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

\[x^{2} + 2x + 1 \leq 3|x + 1|\]

\[(x + 1)^{2} \leq 3|x + 1|\]

\[y = |x + 1|:\]

\[y^{2} \leq 3y\]

\[y^{2} - 3y \leq 0\]

\[y(y - 3) \leq 0\]

\[0 \leq y \leq 3.\]

\[0 \leq |x + 1| \leq 3\]

\[0 \leq |x + 1| - при\ всех\ \text{x.}\]

\[|x + 1| \leq 3\]

\[- 3 \leq x + 1 \leq 3\]

\[- 4 \leq x \leq 2\]

\[Ответ:x \in \lbrack - 4;2\rbrack.\]

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

\[\left( x^{2} + 2x + 2 \right)\left( x^{2} - 4x + 5 \right) = 1\]

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

\[при\ x = - 1;\]

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

\[при\ x = 2;\]

\[Получаем,\ что:\]

\[\left( x^{2} + 2x + 2 \right)\left( x^{2} - 4x + 5 \right) > 1\]

\[Данное\ равенство\ не\ \]

\[выполняется,\ уравнение\ не\ \]

\[имеет\ корней.\]

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

\[\frac{3}{x^{2} + 8x + 17} + \frac{4}{x^{2} + 8x + 18} \geq 5\]

\[y = x^{2} + 8x + 17 =\]

\[= (x + 4)^{2} + 1 \geq 1.\]

\[\frac{3}{y} + \frac{4}{y + 1} \geq 5\ \ \ \ \ \ \ \ | \cdot y(y + 1)\]

\[3(y + 1) + 4y \geq 5\left( y^{2} + y \right)\]

\[3y + 3 + 4y \geq 5y^{2} + 5y\]

\[5y^{2} - 2y - 3 \leq 0\]

\[D = 4 + 60 = 64\]

\[y_{1} = \frac{2 + 8}{10} = 1;\ \ \]

\[y_{2} = \frac{2 - 8}{10} = - 0,6\]

\[5(y + 0,6)(y - 1) \leq 0\]

\[y \in \lbrack - 0,6;1\rbrack\text{.\ }\]

\[Так\ как\ y \geq 1;то\ \]

\[получаем:y = 1.\]

\[Подставим:\]

\[(x + 4)^{2} + 1 = 1\]

\[(x + 4)^{2} = 0\]

\[x + 4 = 0\]

\[x = - 4.\]

\[Ответ:x = - 4.\]