S1 2009 høst LØSNING

1)

$5a^2+(a-2)(a+2)-(2a+1)=5a^2+a^2-4-2a-1=6a^2-2a-5$

2)

$\frac{2(x-2)}{3x}+\frac{1}{2}=\frac{2}{2}\cdot\frac{2x-4}{3x}+\frac{3x}{6x}=\frac{7x-8}{6x}=\frac{7}{6}-\frac{4}{3x}$

3)

$\frac{a^4 \cdot 2b^2}{(2a)^3}=\frac{a^4\cdot2b^2}{2^3\cdot a^3}=\frac{ab^2}{4}$

4)

$lg(a^2b)-lg(\frac{1}{ab})=2lga+lgb-(lg1-lga-lgb)=2lga+lgb-lg1+lga+lgb=3lga+2lgb=lg(a^3b^2)$

1)

$\frac{x}{4}-\frac{1}{6}=\frac{x}{6}-(\frac{x}{2}-1)=>\frac{x}{4}-\frac{x}{6}+\frac{x}{2}=1+\frac{1}{6}$

$x(\frac{1}{4}-\frac{1}{6}+\frac{1}{2})=x\cdot \frac{7}{12}=\frac{7}{6}=>x=2$

2)

$2x^2-2(x+2)=20=>2x^2-2x-24=0$

$2(x^2-x-12)=0=>2(x-4)(x+3)=0=>x=4 \vee x=-3$

c)

masse til skrue målt i gram: s. masse til mutter målt i gram:m

$3s+2m=57$

$ s+3m=33|\cdot3=>3s+9m=99$

$3s+9m-(3s+2m)=99-57=42=>7m=42=>m=6=>s=33-3\cdot6=15$

En skrue veier 15 gram og en mutter veier 6 gram.

d)

$x^2-2x \leq 3=>x^2-2x-3=(x+1)(x-3)\leq0$

$ x\in[-1,3]$