Question

Даю 30 баллов желательно фотографию

Answer 1

$1) \\ \frac{3}{x^2+2}=\frac{1}{x}, x\neq 0\\\\3x=x^2+2\\\\3x-x^2-2=0\\\\-x^2+3x-2=0\\\\x^2-3x+2=0\\\\D=(-3)^2-4\cdot 1\cdot 2 = 9-8 = 1, \; 2 \; kornya\\\\x_{1,2}=\frac{3\pm \sqrt{1}}{2} = \frac{3\pm 1}{2}\\\\x_1 = \frac{3+1}{2}= \frac{4}{2} = 2\\\\x_2 = \frac{3-1}{2} = \frac{2}{2} = 1\\\\Otvet: x_1=1, x_2=2.$

$2)\\ \frac{x^2}{x^2-4} = \frac{5x-6}{x^2-4}, \; x\neq 2, x\neq -2\\\\x^2=5x-6\\\\x^2-5x+6=0\\\\D = (-5)^2 - 4\cdot 1\cdot 6 = 25 - 24 = 1, \; 2 \; kornya\\\\x_{1,2} = \frac{5\pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}\\\\x_1 =\frac{5+1}{2} = \frac{6}{2} = 3\\\\x_2 = \frac{5-1}{2} = \frac{4}{2} = 2\\\\Otvet: \; x = 3.$

$3) \\ \frac{x^2- x-12}{x+3}= 0, \; x \neq -3\\\\\frac{x^2+3x-4x-12}{x-3}=0\\\\\frac{x\cdot (x+3)-4\cdot (x+3)}{x+3}=0\\\\\frac{(x+3)\cdot (x-4)}{x+3}=0\\\\x-4=0\\\\x=4\\\\Otvet:\; x=4$

$4) \\ \frac{x+2}{x-1}+\frac{x}{x+1}=\frac{6}{x^2-1}, \; x\neq 1, \; x\neq -1\\\\\frac{x+2}{x-1}+\frac{x}{x+1}-\frac{6}{x^2-1}=0\\\\\frac{x+2}{x-1}+\frac{x}{x+1}-\frac{6}{(x-1)\cdot (x+1)}=0\\\\\frac{(x+1)\cdot (x+2)+x\cdot (x-1)-6}{(x-1)\cdot (x+1)}=0\\\\\frac{x^2+2x+x+2+x^2-x-6}{(x-1)\cdot (x+1)}=0$

$\frac{2x^2+2x-4}{(x-1)\cdot (x+1)}=0\\\\\frac{2\cdot (x^2+x-2)}{(x-1) \cdot (x+1)}=0\\\\\frac{2\cdot (x^2+2x-x-2)}{(x-1)\cdot (x+1)}=0\\\\\frac{2(x\cdot (x+2)-(x+2)}{(x-1)(x+1)}=0\\\\\frac{2(x+2)(x-1)}{(x-1)(x+1)}=0$

$\frac{2(x+2)}{x+1}=0\\\\2(x+2)=0\\\\x+2=0\\\\x=-2\\\\Otvet: \; x=-2$