Решебник по алгебре 9 класс Рурукин контрольные работы КР-3. Уравнения и неравенства с одной переменной Вариант 4

Условие:

1. Решите уравнение (x^2-12x+20)^2=(x^2+2x-12)^2.

2. Найдите решения уравнения (4x-3)/(3x-4)+(3x-4)/(4x-3)-2=0.

3. Найдите корни уравнения (x^2-25) √(x+3)=0.

4. Решите неравенство (x-1)(2x^2-3x+1)(x+5)≥0.

5. Определите значения a, при которых уравнение 4x³+4x²+ax=0 имеет два корня. Найдите эти корни.

6. Решите неравенство (x^2-5)(4x^2-x-5)<(x^2-3)(4x^2-x-5).

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

\[\left( x^{2} - 12x + 20 \right)^{2} = \left( x^{2} + 2x - 12 \right)^{2}\]

\[1)\ x^{2} - 12x + 20 = x^{2} + 2x - 12\]

\[- 14x + 32 = 0\]

\[- 2(7x - 16) = 0\]

\[7x = 16\]

\[x = \frac{16}{7}.\]

\[2)\ x^{2} - 12x + 20 = - x^{2} - 2x + 12\]

\[2x^{2} - 10x - 8 = 0\ \]

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

\[2(x - 4)(x - 1) = 0\]

\[x = 4;\ \ x = 1\]

\[Ответ:x = \frac{16}{7};\ \ x = 1;\ \ x = 4.\]

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

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

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

\[x + 1 = 0\]

\[x = - 1.\]

\[Ответ:x = - 1.\]

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

\[\left( x^{2} - 25 \right)\sqrt{x + 3} = 0\]

\[x^{2} - 25 = 0\]

\[x^{2} = 25\]

\[x = \pm 5.\]

\[x + 3 \geq 0\]

\[x \geq - 3\]

\[Ответ:x = - 3;\ \ x = 5.\ \]

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

\[(x - 1)\left( 2x^{2} - 3x + 1 \right)(x + 5) \geq 0\]

\[2x^{2} - 3x + 1 = 2 \cdot (x - 0,5)(x - 1)\]

\[D = 9 - 8 = 1\]

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

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

\[(x + 5)(x - 0,5)(x - 1)(x - 1) \geq 0\]

\[Ответ:\]

\[x \in ( - \infty; - 5\rbrack \cup \lbrack 0,5; + \infty).\]

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

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

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

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

\[D = 16 - 16a\]

\[16 - 16a = 0\]

\[16a = 16\]

\[a = 1.\]

\[При\ a = 0:\]

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

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

\[x = 0;\ \ x = - 1.\]

\[При\ a = 1:\]

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

\[x\left( 4x^{2} + 4x + 1 \right) = 0\]

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

\[x = 0;\ \ 2x + 1 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - 0,5.\]

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

\[- 8x^{2} + 2x + 10 < 0\]

\[D = 4 + 320 = 324\]

\[x_{1} = \frac{- 2 + 18}{- 16} = - 1;\ \ \]

\[x_{2} = \frac{- 2 - 18}{- 16} = 1\frac{1}{4}\]

\[(x + 1)(x - 1,25) < 0\]

\[Ответ:\]

\[x \in ( - \infty; - 1) \cup (1,25; + \infty).\]