S1 2018 vår LØSNING

DEL1

Oppgave 1

a)

$2x^2-5x+1=x-3 \\ 2x^2-5x-x+1+3 = 0 \\ 2x^2-6x+4=0 \quad |:2 \\ x^2-3x+2=0$

Bruker abc-formelen $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, $a=1$, $b=-3$, $c=2$.

$x=\frac{-(-3)\pm\sqrt{(-3)^2-4\cdot 1 \cdot 2}}{2\cdot1} \\ x=\frac{3\pm\sqrt{1}}{2} \\ x_1=\frac{3-1}{2} \vee x_2=\frac{3+1}{2} \\ x_1=1 \vee x_2=2$

b)

$2lg(x+7)=4 \quad |:2\\ lg(x+7)=2 \\ 10^{lg(x+7)}=10^2 \\ x+7 = 100 \\ x=93$

c)

$3\cdot2^{3x+2}=12\cdot2^6 \quad |:3 \\ 2^{3x+2} = 4\cdot 2^6 \quad |:2^6 \\ \frac{2^{3x+2}}{2^6} = 4 \\ 2^{3x+2-6}=4 \\ 2^{3x-4}=2^2 \\ 3x-4=2 \\ 3x=6 \\ x=2$

Oppgave 2

<math> \left[ \begin{align*} x^2 + 3y = 7 \\ 3x - y = 1 \end{align*}\right] </math>

Løser likning to med hensyn på y:

$3x-y=1 \\ 3x-1=y \\ y=3x-1$

Bruker innsettingsmetoden og erstatter y med 3x-1 i likning én.

$x^2+3(3x-1) = 7 \\ x^2+9x-3-7=0 \\ x^2 +9x-10=0$

Bruker abc-formelen $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, $a=1$, $b=9$, $c=-10$.

$x=\frac{-9\pm\sqrt{(9)^2-4\cdot 1 \cdot (-10)}}{2\cdot1} \\ x=\frac{-9\pm\sqrt{121}}{2} \\ x_1=\frac{-9-11}{2} \vee x_2=\frac{-9+11}{2} \\ x_1=-10 \vee x_2=1$

Bruker likning to for å finne tilhørende y-verdier:

$y=3x-1 \\ y_1=3\cdot (-10)-1 \quad \vee \quad y_2=3\cdot 1 - 1 \\ y_1=-31 \quad \vee \quad y_2=2$

Løsning: $x_1=-10 \wedge y_1=-31 \quad \vee \quad x_2=1 \wedge y_2=2 $

Oppgave 3

a)

$(2x-3)^2-2x(2x-6)\\=(2x)^2-2\cdot2x\cdot3+3^2-2x\cdot2x-2x\cdot(-6)\\=4x^2-12x+9-4x^2+12x\\=9$

b)

$lg(2a)+lg(4a)+lg(8a)-lg(16a)\\=lg(\frac{2a\cdot4a\cdot8a}{16a})\\=lg(4a^2)\\=lg(2a)^2\\=2lg(2a)$

c)

$\frac{1}{a}+\frac{1}{b}-\frac{a-b}{ab}\\=\frac{1\cdot b}{a\cdot b}+\frac{1\cdot a}{a\cdot b}-\frac{a-b}{ab}\\=\frac{b}{ab}+\frac{a}{ab}-\frac{a-b}{ab}\\=\frac{b+a-a+b}{ab}\\=\frac{2b}{ab}\\=\frac{2}{a}$

Oppgave 4

Løser likningen

$x^2-3x+2=0$