DEL 1

Oppgave 1

a)

$x^{2} - 3 x + 1 = 3 x + 8 x^{2} - 6 x - 7 = 0 x = \frac{6 \pm \sqrt{(- 6)^{2} - 4 \cdot (- 7)}}{2} x = \frac{6 \pm 8}{2} x_{1} = - 1 \lor x_{2} = 7$

b)

$l g (x^{4}) - l g (x^{3}) + l g (x^{2}) - l g x = 6 4 l g x - 3 l g x + 2 l g x - l g x = 6 2 l g x = 6 l g x = 3 x = 10^{3} x = 1000$

c)

$10 \cdot 4^{x} = 5 \cdot 2^{x} \frac{2^{2 x}}{2^{x}} = \frac{5}{10} 2^{2 x - x} = \frac{1}{2} 2^{x} = 2^{- 1} x = - 1$

Oppgave 2

a)

$(a + 2 b)^{2} - (2 b - a)^{2} = (a^{2} + 4 a b + 4 b^{2}) - (4 b^{2} - 4 a b + a^{2}) = a^{2} + 4 a b + 4 b^{2} - 4 b^{2} + 4 a b - a^{2} = 8 a b$

b)

$3^{(} 3) \cdot 3^{(} 0) + 3^{- 1} + 3^{- 2} + 3^{- 3} = 27 \cdot 1 + \frac{1}{3} + \frac{1}{3^{(} 2)} + \frac{1}{3^{(} 3)} = 27 + \frac{9}{27} + \frac{3}{27} + \frac{1}{27} = 27 + \frac{13}{27} = \frac{729}{27} + \frac{13}{27} = \frac{742}{27}$

S1 2018 høst LØSNING

DEL 1

Oppgave 1

a)

b)

c)

Oppgave 2

a)

b)

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