Odpowiedzi

2010-04-09T15:30:26+02:00
Obliczylem zaznaczone przykaldy,bo chyba o to chodzilo

pierwszy:

x-1/x²+x+1*x³-1/x²√2-√2*(1+1/x-1-2):2-x/√2=
(x-1/x²+x+1)*(x-1)(x²+x+1)/√(x-1)(x+1)*(x-1+1-2x+2/x-1)*√2/2-x=x-1/√2(x+1)*(2-x/x-1)*√2/2-x=1/x+1

drugi:

(x/x³-1-1/x²):(x/x²+x+1+1-x/x²)=
(x³-x³+1/(x³-1)*x²):(x³+x²-x³+x-x²+1-x)/x²(x²+x+1)=
1/(x-1)(x²+x+1)*x²(x²+x+1)/1=
x²/x-1