Question

1. найти сумму корней уравнения 2. найти сумму целых решений неравенства 3. указать количество корней уравнения из промежутка

Answer 1

1.
$\sqrt{x+1}- \sqrt{9-x} = \sqrt{2x-12}, \\ \begin{cases}x+1 \geq 0,\\9-x \geq 0,\\2x-12 \geq 0;\end{cases} \ \begin{cases}x \geq -1,\\x \leq 9,\\x \geq 6;\end{cases} \ 6 \leq x \leq 9; \\$
$( \sqrt{x+1}- \sqrt{9-x} )^2 = (\sqrt{2x-12})^2, \\ (\sqrt{x+1})^2- 2\sqrt{x+1}\cdot\sqrt{9-x}+(\sqrt{9-x})^2 = 2x-12, \\ x+1 - 2\sqrt{(x+1)(9-x)}+9-x = 2x-12, \\ - 2\sqrt{-x^2+8x+9} = 2x-22, \\ \sqrt{-x^2+8x+9} = 11-x, \\ (\sqrt{-x^2+8x+9})^2 = (11-x)^2, \\ -x^2+8x+9 = 121-22x+x^2, \\ 2x^2-30x+112=0 , \\ x^2-15x+56=0, \\ x_1+x_2=15.$
2.
$3x-|6x-18|0, \\ |6x-18|<3x, \\ x \geq 0, \\ \left \{ {{6x-18<3x,} \atop {6x-18-3x;}} \right. \ \left \{ {{3x<18,} \atop {9x18;}} \right. \ \left \{ {{x<6,} \atop {x2;}} \right. \ 2<x<6; \\x\in(2;6); \\ 3+4+5=12.$
3.
$\sin2x = \sqrt2\cos(\frac{\pi}{2}+x), \\ 2\sin x\cos x = -\sqrt2\sin x, \\ 2\sin x\cos x + \sqrt2\sin x = 0, \\ \sin x(2\cos x + \sqrt2) = 0, \\ \left \ [ {{\sin x = 0,} \atop {\cos x = -\frac{\sqrt2}{2};}} \right. \ \left \ [ {{x=\pi k, k\in Z,} \atop {x = \pm\frac{3\pi}{4}+2\pi k, k\in Z;}} \right. \\$
$x\in[-2\pi;-\pi], \\ \left[\begin{array} \ -2\pi \leq \pi k \leq -\pi, \\-2\pi \leq \frac{3\pi}{4}+2\pi k \leq -\pi, \\-2\pi \leq -\frac{3\pi}{4}+2\pi k \leq -\pi; \end{array} \ \left[\begin{array} \ -2 \leq k \leq -1, \\ - 1\frac{3}{8} \leq k \leq - \frac{7}{8} , \\ -\frac{5}{8} \leq k \leq -\frac{1}{8}; \end{array} \ \ \left[\begin{array} \ k\in \{-2;-1\}, \\k=-1 , \\ k\in\varnothing; \end{array}$
3 корня.