1T 2010 høst LØSNING: Forskjell mellom sideversjoner
Fra Matematikk.net
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d) | d) | ||
<p></p> | <p></p> | ||
<tex> | |||
\frac{3}{x+4} + \frac{24}{x^2-16} = \frac{3}{x+4} + \frac{24}{(x+4)(x-4)}= \frac{3 (x-4)}{(x+4)(x-4)} + \frac{24}{(x+4)(x-4)}= \\ | |||
\frac{3x-12}{(x+4)(x-4)} + \frac{24}{(x+4)(x-4)}= \frac{3x + 12}{(x+4)(x-4)} = | |||
</tex> | |||
== Oppgave 2 == | == Oppgave 2 == | ||
Sideversjonen fra 17. feb. 2012 kl. 06:20
DEL 1.
Oppgave 1
a)
<tex> x+y=4 \wedge 3x - y =8</tex>
<tex>y=4-x \wedge 3x-4+x=8</tex>
<tex>y=4-x \wedge 4x=12</tex>
<tex>y=4-x \wedge x=3</tex>
<tex>y=1 \wedge x=3</tex>
<tex>x=3 \wedge y=1</tex>
b)
<tex>- \frac14x+2 =2x- \frac52</tex>
<tex>x+8 =8x- 10</tex>
<tex>- 9x = - 18</tex>
<tex>x = 2</tex>
c)
<tex>5,7 \cdot 10^4 + 3,0 \cdot 10^3 = 57000 + 3000 = 60000 = 6,0 \cdot 10^4</tex>
d)
<tex> \frac{3}{x+4} + \frac{24}{x^2-16} = \frac{3}{x+4} + \frac{24}{(x+4)(x-4)}= \frac{3 (x-4)}{(x+4)(x-4)} + \frac{24}{(x+4)(x-4)}= \\ \frac{3x-12}{(x+4)(x-4)} + \frac{24}{(x+4)(x-4)}= \frac{3x + 12}{(x+4)(x-4)} = </tex>
Oppgave 2
DEL 2