Question

Найдите количество целых корней уравнения |x^2-7x-8|: (x^2-7x-8)=|x^2+2x-8|: (-x^2-2x+8)

Answer 1

$x^2-7x-8 = 0, \\ x_1=-1, x_2=8; \\ x^2+2x-8=0, \\ x_1=-4, x_2=2; \\ \\ x\ \textless \ -4, \\ x^2-7x-8\ \textgreater \ 0, x^2+2x-8\ \textgreater \ 0, \\ (x^2-7x-8):(x^2-7x-8)=(x^2+2x-8):(-x^2-2x+8), \\ 1=-1, \\ 0\cdot x = -2, \\ x\in\varnothing; \\ \\ -4\ \textless \ x\ \textless \ -1, \\ x^2-7x-8\ \textgreater \ 0, x^2+2x-8\ \textless \ 0, \\ (x^2-7x-8):(x^2-7x-8)=-(x^2+2x-8):(-x^2-2x+8), \\ 1=1, \\ 0\cdot x = 0, \\ x\in R, \\ x\in(-4;-1); \\ \\ -1\ \textless \ x\ \textless \ 2, \\ x^2-7x-8\ \textless \ 0, x^2+2x-8\ \textless \ 0, \\ -(x^2-7x-8):(x^2-7x-8)=-(x^2+2x-8):(-x^2-2x+8), \\ -1=1, \\ 0\cdot x = 2, \\ x\in\varnothing; \\ \\$

$2\ \textless \ x\ \textless \ 8, \\ x^2-7x-8\ \textless \ 0, x^2+2x-8\ \textgreater \ 0, \\ -(x^2-7x-8):(x^2-7x-8)=(x^2+2x-8):(-x^2-2x+8), \\ -1=-1, \\ 0\cdot x = 0, \\ x\in R; \\ x\in(2;8); \\ \\ x\ \textgreater \ 8, \\ x^2-7x-8\ \textgreater \ 0, x^2+2x-8\ \textgreater \ 0, \\ (x^2-7x-8):(x^2-7x-8)=(x^2+2x-8):(-x^2-2x+8), \\ 1=-1, \\ 0\cdot x = -2, \\ x\in\varnothing; \\ \\ x\in(-4;-1)\cup(2;8); \\ X_Z=\{-3, -2, 3, 4, 5, 6, 7\}, \\ 7.$