1T 2019 vår LØSNING: Forskjell mellom sideversjoner

Sideversjonen fra 7. aug. 2019 kl. 04:41

$\frac{4,5 \cdot 10^{12}}{900} = \frac{4,5}{9} \cdot \frac{10^{12}}{10^2} = 0,5 \cdot 10^{10} = 5,0 \cdot 10^9$

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

@@ Linje 19: / Linje 19: @@
 ===Oppgave 3===
-$\frac{x^2}{x^2-4} + \frac{3}{x-2} + \frac{1}{x+2} \\ = \frac{x^2}{(x+2)(x-2)} + \frac {3 (x+2)}{(x+2)(x-2)} +$ \frac{1  <span style="color:red">(x-2)</span>}{(x+2)(x-2)}$
+$\frac{x^2}{x^2-4} + \frac{3}{x-2} + \frac{1}{x+2} \\ = \frac{x^2}{(x+2)(x-2)} + \frac {3 (x+2)}{(x+2)(x-2)} +\frac{1 (x-2)}{(x+2)(x-2)}$
 ===Oppgave 4===