вправа 23.96 гдз 11 клас алгебра Істер 2019
Вправа 23.96
Умова:
Розв'яжіть нерівність:
Розв'яжіть нерівність:
Відповідь ГДЗ:
\begin{equation}
1)\sqrt{36-6x-6x^{2}}<x+6
\end{equation}
\begin{equation}
\left\{\begin{matrix}
36-6x-6x^{2}\geq 0 & & \\
x+6>0 & & \\
36-6x-6x^{2}<(x+6)^{2} & &
\end{matrix}\right.
\end{equation}
x + 6 > 0, x > -6
36 - 6x - 6x2 ≥ 0
-x2 - x + 6 ≥ 0
Д = (-1)2 - 4 • (-1) • 6 = 25 \begin{equation} x_{1;2}=\frac{1\pm 5}{-2}=-3;2 \end{equation} 36 - 6x - 6x2 < x2 + 12x + 36
-6x2 - x2 - 6x - 12x < 0
-7x2 - 18x < 0
x(-7x - 18) < 0
x1 = 0
-7x - 18 = 0
-7x = 18 \begin{equation} x=-\frac{18}{7} \end{equation}
\begin{equation} x\in [-3;-\frac{18}{7})U(0;2] \end{equation} \begin{equation} 2)\sqrt{x^{2}-5x+6}\leq 3x-6 \end{equation} \begin{equation} \left\{\begin{matrix} x^{2}-5x+6\geq 0 & & \\ 3x-6>0 & & \\ (\sqrt{x^{2}-5x+6})^{2}\leq (3x-6)^{2} & & \end{matrix}\right. \end{equation} 3x - 6 > 0
x > 2
x2 - 5x + 6 ≥ 0
Д = 1 \begin{equation} x_{1;2}=\frac{5\pm 1}{2}=3;2 \end{equation} x2 - 5x + 6 ≤ 9x2 - 36x + 36
x2 - 9x2 - 5x + 36 + 6 - 36 ≤ 0
-8x2 + 31x - 30 ≤ 0
Д = 312 - 4 • (-8) • (-30) =
= 961 - 960 = 1 \begin{equation} x_{1;2}=\frac{-31\pm 1}{-16}=2;\frac{15}{8} \end{equation}
х ∈ {2} U [3; +∞). \begin{equation} 3)\sqrt{x^{2}+6x-16}>-x-4 \end{equation} \begin{equation} \left\{\begin{matrix} x^{2}+6x-16\geq 0 & & \\ -x-4>0 & & \\ x^{2}+6x-16>(-(x+4))^{2} & & \end{matrix}\right. \end{equation} -x - 4 > 0
-x > 4 • (-1)
x < -4
x2 + 6x - 16 ≥ 0
Д = 36 + 64 = 100 \begin{equation} x_{1;2}=\frac{-6\pm 10}{2}=-8;2 \end{equation} x2 + 6x - 16 > x2 + 8x + 16
6x - 8x > 32
-2x > 32 • (-1)
2x < 32
x < -16
х ∈ (-∞; -16). \begin{equation} 4)\sqrt{2x^{2}-4x-7}>x-1 \end{equation} \begin{equation} \left\{\begin{matrix} 2x^{2}-4x-7\geq 0 & & \\ x-1>0 & & \\ 2x^{2}-4x-7>(x-1)^{2} & & \end{matrix}\right. \end{equation} x - 1 > 0, x > 1
2x2 - 4x - 7 ≥ 0
Д = (-4)2 - 4 • (-7) • 2 = 72 \begin{equation} x_{1;2}=\frac{4\pm 6\sqrt{2}}{4} \end{equation} \begin{equation} x_{1;2}=\frac{2(2\pm 3\sqrt{2})}{4}= \end{equation} \begin{equation} =\frac{2\pm 3\sqrt{2})}{2} \end{equation} 2x2 - 4x - 7 > x2 - 2x + 1
2x2 - x2 - 4x + 2x - 7 - 1 > 0
x2 - 2x - 8 > 0
Д = 4 - 4 • (-8) = 36 \begin{equation} x_{1;2}=\frac{2\pm 6}{2}=4;-2 \end{equation}
\begin{equation} x\in (-\propto ;\frac{2-3\sqrt{2}}{2}]U(4;+\propto ) \end{equation}
36 - 6x - 6x2 ≥ 0
-x2 - x + 6 ≥ 0
Д = (-1)2 - 4 • (-1) • 6 = 25 \begin{equation} x_{1;2}=\frac{1\pm 5}{-2}=-3;2 \end{equation} 36 - 6x - 6x2 < x2 + 12x + 36
-6x2 - x2 - 6x - 12x < 0
-7x2 - 18x < 0
x(-7x - 18) < 0
x1 = 0
-7x - 18 = 0
-7x = 18 \begin{equation} x=-\frac{18}{7} \end{equation}
\begin{equation} x\in [-3;-\frac{18}{7})U(0;2] \end{equation} \begin{equation} 2)\sqrt{x^{2}-5x+6}\leq 3x-6 \end{equation} \begin{equation} \left\{\begin{matrix} x^{2}-5x+6\geq 0 & & \\ 3x-6>0 & & \\ (\sqrt{x^{2}-5x+6})^{2}\leq (3x-6)^{2} & & \end{matrix}\right. \end{equation} 3x - 6 > 0
x > 2
x2 - 5x + 6 ≥ 0
Д = 1 \begin{equation} x_{1;2}=\frac{5\pm 1}{2}=3;2 \end{equation} x2 - 5x + 6 ≤ 9x2 - 36x + 36
x2 - 9x2 - 5x + 36 + 6 - 36 ≤ 0
-8x2 + 31x - 30 ≤ 0
Д = 312 - 4 • (-8) • (-30) =
= 961 - 960 = 1 \begin{equation} x_{1;2}=\frac{-31\pm 1}{-16}=2;\frac{15}{8} \end{equation}
х ∈ {2} U [3; +∞). \begin{equation} 3)\sqrt{x^{2}+6x-16}>-x-4 \end{equation} \begin{equation} \left\{\begin{matrix} x^{2}+6x-16\geq 0 & & \\ -x-4>0 & & \\ x^{2}+6x-16>(-(x+4))^{2} & & \end{matrix}\right. \end{equation} -x - 4 > 0
-x > 4 • (-1)
x < -4
x2 + 6x - 16 ≥ 0
Д = 36 + 64 = 100 \begin{equation} x_{1;2}=\frac{-6\pm 10}{2}=-8;2 \end{equation} x2 + 6x - 16 > x2 + 8x + 16
6x - 8x > 32
-2x > 32 • (-1)
2x < 32
x < -16
х ∈ (-∞; -16). \begin{equation} 4)\sqrt{2x^{2}-4x-7}>x-1 \end{equation} \begin{equation} \left\{\begin{matrix} 2x^{2}-4x-7\geq 0 & & \\ x-1>0 & & \\ 2x^{2}-4x-7>(x-1)^{2} & & \end{matrix}\right. \end{equation} x - 1 > 0, x > 1
2x2 - 4x - 7 ≥ 0
Д = (-4)2 - 4 • (-7) • 2 = 72 \begin{equation} x_{1;2}=\frac{4\pm 6\sqrt{2}}{4} \end{equation} \begin{equation} x_{1;2}=\frac{2(2\pm 3\sqrt{2})}{4}= \end{equation} \begin{equation} =\frac{2\pm 3\sqrt{2})}{2} \end{equation} 2x2 - 4x - 7 > x2 - 2x + 1
2x2 - x2 - 4x + 2x - 7 - 1 > 0
x2 - 2x - 8 > 0
Д = 4 - 4 • (-8) = 36 \begin{equation} x_{1;2}=\frac{2\pm 6}{2}=4;-2 \end{equation}
\begin{equation} x\in (-\propto ;\frac{2-3\sqrt{2}}{2}]U(4;+\propto ) \end{equation}