Question

$( \frac{1}{ x^{2} -2x} - \frac{1}{ x^{2}+2x } + \frac{2}{ x^{2} +4} ): \frac{2x+6}{ x^{3} -4x}$

Answer 1

[tex]E(x)=(\frac{1}{x^2-2x}-\frac{1}{x^2+2x}+\frac{2}{x^2-4}):\frac{2x+6}{x^3-4x}\\ E(x)=[\frac{1}{x(x-2)}-\frac{1}{x(x+2)}+\frac{2}{(x-2)(x+2)}]*\frac{x(x^2-4)}{2(x+3)}\\ E(x)=\frac{x+2-x+2+2x}{x(x-2)(x+2)}*\frac{x(x-2)(x+2)}{2(x+3)}\\ E(x)=\frac{2x+4}{x(x-2)(x+2)}*\frac{x(x-2)(x+2)}{2(x+3)}\\ Se\ simplifica\ cativa\ termeni\ si\ ramane:\\ E(x)=\frac{2x+4}{2(x+3)}\\ E(x)=\frac{2(x+2)}{2(x+3)}\\ E(x)=\frac{x+2}{x+3}[/tex]

Answer 2

[1/(x²-2x) -1/(x²+2x)+2/(x²-4)] : (2x+6)/(x³-4x)=
=[1/x(x-2) -1/x(x+2)+2/(x²-4)] : 2(x+3)/x(x²-4)=
=[(x+2-x+2+2x/x(x-2)(x+2)+2/(x²-4)] : 2(x+3)/x(x²-4)=
=[(4+2x)/x(x²-4)] : 2(x+3)/x(x²-4)=
=2(2+x)/x(x²-4) × x(x²-4)/2(x+3)=
=(x+2)/(x+3)