Question

dau cel mai bun raspuns

Answer 1

 [tex] Partea \,1.\\
\frac{1}{x(x-1)}-\frac{1}{x(x+1)}+\frac{x+3}{(x+1)(x-1)}=\\
=\frac{x+1-x+1+x^2+3x}{x(x+1)(x-1)}=\frac{x^2-3x+2}{x(x+1)(x-1)}=\\ =\frac{x^2-x-2x+2}{x(x-1)(x+1)}=\frac{(x-1)(x-2)}{x(x-1)(x+1)}=\\
=\frac{(x-2)}{x(x+1)}\;\surd\surd\;\\
Partea\;2.\; Numar\u{a}torul\;=x(x^2+4x-5)=\\
=x(x^2-x+5x-5)=x(x-1)(x+5)\surd\;\\
numitorul\,=x^2-x-6=x^2-3x+2x-6=\\ =x(x-3)+2(x-3)=(x+2)(x-3)\;\surd\\
P_1\;\cdot\;P_2= ... de\,aici\;ar\;trebuii\;sa\;te\;descurci\;,\,succes![/tex]