Question

Даю 30 баллов желательно фотографию

Answer 1

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

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

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

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

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

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