ГДЗ по алгебре 9 класс Рурукин контрольные работы КР-3. Уравнения и неравенства с одной переменной Вариант 6

Условие:

1. Решите уравнение x^4+x^3-2x^2-5x-2=0.

2. Найдите решения уравнения (x+3x/(x-3))^2=4-(3x^2)/(x-3).

3. Найдите корни уравнения x(x+1)(x+2)(x+3)=120.

4. Решите неравенство x^2-5|x-5|-10x≤-25.

5. Докажите, что уравнение (x²-2x+3)(x²-6x+10)=2 не имеет корней.

6. Решите неравенство 2/(x^2+10x+27)+5/(x^2+10x+26)≥6.

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

\[x^{4} + x^{3} - 2x^{2} - 5x - 2 = 0\]

\[x = - 1:\]

\[разделим\ на\ (x + 1)\ и\ получим\]

\[x^{3} - 3x - 2 = 0.\]

\[x = 2:\]

\[разделим\ на\ (x - 2)\ и\ получим\]

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

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

\[x + 1 = 0\]

\[x = - 1.\]

\[Ответ:x = - 1;x = 2.\]

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

\[\left( x^{\backslash x - 3} + \frac{3x}{x - 3} \right)^{2} = 4 - \frac{3x^{2}}{x - 3}\]

\[\left( \frac{x^{2} - 3x + 3x}{x - 3} \right)^{2} = 4 - 3\frac{x^{2}}{x - 3}\]

\[\left( \frac{x^{2}}{x - 3} \right)^{2} = 4 - 3\frac{x^{2}}{x - 3}\]

\[y = \frac{x^{2}}{x - 3}:\]

\[y^{2} = 4 - 3y\]

\[y^{2} + 3y - 4 = 0\]

\[D = 9 + 16 = 25\]

\[y_{1} = \frac{- 3 + 5}{2} = 1;\ \ \]

\[y_{2} = \frac{- 3 - 5}{2} = - 4\]

\[1)\ y = - 4:\]

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

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

\[D = 4 + 12 = 16\]

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

\[x_{2} = - 2 - 4 = - 6.\]

\[2)\ y = 1:\]

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

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

\[D = 1 - 12 < 0\]

\[нет\ корней.\]

\[Ответ:x = 2;x = - 6.\]

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

\[x(x + 1)(x + 2)(x + 3) = 120\]

\[\left( x(x + 3) \right)\left( (x + 1)(x + 2) \right) = 120\]

\[\left( x^{2} + 3x \right)\left( x^{2} + 3x + 2 \right) = 120\]

\[y = x^{2} + 3x:\]

\[y(y + 2) = 120\]

\[y^{2} + 2y - 120 = 0\]

\[D = 1 + 120 = 121\]

\[y_{1} = - 1 + 11 = 10;\ \ \]

\[y_{2} = - 1 - 11 = - 12\]

\[1)\ y = - 12:\]

\[x^{2} + 3x = - 12\]

\[x^{2} + 3x + 12 = 0\]

\[D = 9 - 48 < 0\]

\[нет\ корней.\]

\[2)\ y = 10:\]

\[x^{2} + 3x = 10\]

\[x^{2} + 3x - 10 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 10\]

\[x_{1} = - 5;\ \ x_{2} = 2.\]

\[Ответ:x = - 5;x = 2.\]

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

\[x^{2} - 5|x - 5| - 10x \leq - 25\]

\[x^{2} - 10x + 25 \leq 5|x - 5|\]

\[(x - 5)^{2} \leq 5|x - 5|\]

\[y = |x - 5|:\]

\[y^{2} \leq 5y\]

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

\[y(y - 5) \leq 0\]

\[0 \leq y \leq 5\]

\[0 \leq |x - 5| - верно\ при\ всех\ \text{x.}\]

\[|x - 5| \leq 5\]

\[- 5 \leq x - 5 \leq 5\]

\[0 \leq x \leq 10\]

\[Ответ:x \in \lbrack 0;10\rbrack.\]

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

\[\left( x^{2} - 2x + 3 \right)\left( x^{2} - 6x + 10 \right) = 2\]

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

\[при\ x = 1;\]

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

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

\[при\ x = 3;\]

\[Получаем,\ что\ \]

\[\left( x^{2} - 2x + 3 \right)\left( x^{2} - 6x + 10 \right) \geq 2.\]

\[Значит,\ данное\ равенство\ не\ \]

\[выполняется\ и\ уравнение\ не\ \]

\[имеет\ корней.\]

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

\[\frac{2}{x^{2} + 10x + 27} + \frac{5}{x^{2} + 10x + 26} \geq 6\]

\[y = x^{2} + 10x + 26 =\]

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

\[\frac{2}{y + 1} + \frac{5}{y} \geq 6\ \ \ \ \ \ | \cdot y(y + 1)\]

\[2y + 5 \cdot (y + 1) \geq 6 \cdot \left( y^{2} + y \right)\]

\[2y + 5y + 5 \geq 6y^{2} + 6y\]

\[6y^{2} - y - 5 \leq 0\]

\[D = 1 + 120 = 121\]

\[y_{1} = \frac{1 + 11}{12} = 1;\ \ \]

\[y_{2} = \frac{1 - 11}{12} = - \frac{10}{12} = - \frac{5}{6}\]

\[6\left( y + \frac{5}{6} \right)(y - 1) \leq 0\]

\[y \in \left\lbrack - \frac{5}{6};1 \right\rbrack.\]

\[Так\ как\ y \geq 1;то\ \]

\[получаем:y = 1.\]

\[Подставим:\]

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

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

\[x + 5 = 0\]

\[x = - 5.\]

\[Ответ:x = - 5.\]

## КР-4. Уравнения и неравенства с двумя переменными