Question

48//х - 40//ч+4 =1
решите пожалуйста дробное ур.

Answer 1

$\frac{48}{x}-\frac{40}{x+4}=1;\; \; ODZ:\; x\ne 0,\; x\ne -4\\\\\frac{48(x+4)-40x-x(x+4)}{x(x+4)}=0\\\\\frac{48x+192-40x-x^2-4x}{x(x+4)}=0\\\\\frac{x^2-4x-192}{x(x+4)}=0\\\\x^2-4x-192=0\; \to \; x_1=-12,\; x_2=16\\\\\frac{(x+12)(x-16)}{x(x+4)}=0\\\\x_1=-12,\; x_2=16$

Answer 2

$\frac{48}{x} - \frac{40}{x+4}=1 ; \ \ \ \ ODZ: \left \{ {{x+4 \neq 0} \atop {x \neq 0}} \right. ; \left \{ {{x \neq -4} \atop {x \neq 0}} \right. \\ \frac{48x+192-40x}{x^2+4x}=1\\8x+192=x^2+4x \\-x^2+4x+192=0 \\D=16+768=784\\x_1=(-4+28):(-2)=-12\\x_2=(-4-28):(-2)=16\\ OTVET: x_1=-12 \ \ ; x_2=16$