[( 1+ \frac{x-2} {x} )( \frac{2}{x-1} +2)-\frac{4}{x+1} ]*\frac{3x+3}{2x}=[\frac{2}{x-1}+2+\frac{2(x-2)} {x(x-1)}+\frac{2(x-2)} {x}- \frac{4}{x+1}]*\frac{3x+3}{2x}=
=[\frac{2x}{x(x-1)}+\frac{2x-4} {x(x-1)}+\frac{(2x-4)(x-1)} {x(x-1)}- \frac{4}{x+1}+2]*\frac{3x+3}{2x}=
=(\frac{2x+2x-4+2x^2-2x-4x+4}{x(x-1)}-\frac{4}{x+1}+2)*\frac{3x+3}{2x}=
=\frac{2x^2-2x}{x(x-1)}-\frac{4}{x+1}+2)*\frac{3x+3}{2x}=[\frac{2x(x-1)}{x(x-1)}-\frac{4}{x+1}+2)*\frac{3x+3}{2x}=
=(2- \frac{4}{x+1} +2)*\frac{3x+3}{2x}= (\frac{4(x+1)-4}{x+1})*\frac{3x+3}{2x}=
=(\frac{4x+4-4}{x+1})* \frac{3(x+1)}{2x}=\frac{4x}{x+1}*\frac{3(x+1)}{2x}=6
Czy tak ma być to zapisane?
Mozolne liczenie ale najważniejszy jest rezultat. PRAWDA ???