Oppgaver: Tallregning#
Oppgave 1#
Regn ut.
a)
\[
2 \cdot 3 + 5 - 2\cdot 6
\]
\[
-1
\]
\[\begin{split}
\begin{align*}
2 \cdot 3 + 5 - 2\cdot 6 &= 6 + 5 - 12\\
\\
&= 11 - 12\\
\\
&= -1
\end{align*}
\end{split}\]
b)
\[
4 \cdot 2 - 7 + 3 \cdot 5
\]
\[
16
\]
\[\begin{split}
\begin{align*}
4 \cdot 2 - 7 + 3 \cdot 5 &= 8 - 7 + 15\\
\\
&= 1 + 15\\
\\
&= 16
\end{align*}
\end{split}\]
c)
\[
8 - 3 \cdot 2 + 6 \cdot 2
\]
\[
14
\]
\[\begin{split}
\begin{align*}
8 - 3 \cdot 2 + 6 \cdot 2 &= 8 - 6 + 12\\
\\
&= 2 + 12\\
\\
&= 14
\end{align*}
\end{split}\]
d)
\[
5 \cdot 4 - 2 \cdot 7 + 1
\]
\[
7
\]
\[\begin{split}
\begin{align*}
5 \cdot 4 - 2 \cdot 7 + 1 &= 20 - 14 + 1\\
\\
&= 6 + 1\\
\\
&= 7
\end{align*}
\end{split}\]
Oppgave 2#
Primtallsfaktoriser tallene.
a)
\[
12
\]
\[
12 = 2^2 \cdot 3
\]
\[
12 = 4 \cdot 3 = 2^2 \cdot 3
\]
b)
\[
30
\]
\[
30 = 2 \cdot 3 \cdot 5
\]
\[
30 = 2 \cdot 3 \cdot 5
\]
c)
\[
62
\]
\[
62 = 2 \cdot 31
\]
\[
62 = 2 \cdot 31
\]
d)
\[
102
\]
\[
102 = 2 \cdot 3 \cdot 17
\]
\[
102 = 2 \cdot 3 \cdot 17
\]
Oppgave 3#
Forkort brøkene så mye som mulig.
a)
\[
\dfrac{4}{8}
\]
\[
\dfrac{1}{2}
\]
\[
\dfrac{4}{8} = \dfrac{1\cdot 4}{2\cdot 4} = \dfrac{1}{2}
\]
b)
\[
\dfrac{6}{9}
\]
\[
\dfrac{2}{3}
\]
\[
\dfrac{6}{9} = \dfrac{2\cdot 3}{3\cdot 3} = \dfrac{2}{3}
\]
c)
\[
\dfrac{18}{40}
\]
\[
\dfrac{9}{20}
\]
\[
\dfrac{18}{40} = \dfrac{2\cdot 9}{2\cdot 20} = \dfrac{9}{20}
\]
d)
\[
\dfrac{9}{36}
\]
\[
\dfrac{1}{4}
\]
\[
\dfrac{9}{36} = \dfrac{1\cdot 9}{4\cdot 9} = \dfrac{1}{4}
\]
Oppgave 4#
Regn ut og forkort svaret så mye som mulig.
a)
\[
\dfrac{1}{4} + \dfrac{2}{3}
\]
\[
\dfrac{11}{12}
\]
\[
\dfrac{1}{4} + \dfrac{2}{3} = \dfrac{1\cdot 3}{4\cdot 3} + \dfrac{2 \cdot 4}{3 \cdot 4} = \dfrac{3}{12} + \dfrac{8}{12} = \dfrac{11}{12}
\]
b)
\[
\dfrac{1}{3} - \dfrac{7}{12}
\]
\[
-\dfrac{1}{4}
\]
\[
\dfrac{1}{3} - \dfrac{7}{12} = \dfrac{1\cdot 4}{3\cdot 4} - \dfrac{7}{12} = \dfrac{4}{12} - \dfrac{7}{12} = -\dfrac{3}{12} = -\dfrac{1}{4}
\]
c)
\[
\dfrac{2}{5} + \dfrac{3}{10}
\]
\[
\dfrac{7}{10}
\]
\[
\dfrac{2}{5} + \dfrac{3}{10} = \dfrac{2\cdot 2}{5\cdot 2} + \dfrac{3}{10} = \dfrac{4}{10} + \dfrac{3}{10} = \dfrac{7}{10}
\]
d)
\[
\dfrac{5}{6} - \dfrac{1}{4}
\]
\[
\dfrac{7}{12}
\]
\[
\dfrac{5}{6} - \dfrac{1}{4} = \dfrac{5\cdot 2}{6\cdot 2} - \dfrac{1 \cdot 3}{4 \cdot 3} = \dfrac{10}{12} - \dfrac{3}{12} = \dfrac{7}{12}
\]
Oppgave 5#
Regn ut.
a)
\[
2 \cdot \dfrac{3}{4} + 5 - 4\cdot \dfrac{1}{2}
\]
\[
\dfrac{9}{2}
\]
\[\begin{split}
\begin{align*}
2 \cdot \dfrac{3}{4} + 5 - 4\cdot \dfrac{1}{2} &= \dfrac{6}{4} + 5 - \dfrac{4}{2}\\
\\
&= \dfrac{3}{2} + 5 - 2\\
\\
&= \dfrac{3}{2} + 3\\
\\
&= \dfrac{3}{2} + \dfrac{6}{2}\\
\\
&= \dfrac{9}{2}
\end{align*}
\end{split}\]
b)
\[
3 \cdot \dfrac{5}{6} - 2 + \dfrac{1}{3}
\]
\[
\dfrac{5}{6}
\]
\[\begin{split}
\begin{align*}
3 \cdot \dfrac{5}{6} - 2 + \dfrac{1}{3} &= \dfrac{15}{6} - 2 + \dfrac{1}{3}\\
\\
&= \dfrac{5}{2} - 2 + \dfrac{1}{3}\\
\\
&= \dfrac{15}{6} - \dfrac{12}{6} + \dfrac{2}{6}\\
\\
&= \dfrac{15 - 12 + 2}{6}\\
\\
&= \dfrac{5}{6}
\end{align*}
\end{split}\]
c)
\[
\dfrac{7}{8} + 2 \cdot \dfrac{1}{4} - \dfrac{3}{2}
\]
\[
-\dfrac{1}{8}
\]
\[\begin{split}
\begin{align*}
\dfrac{7}{8} + 2 \cdot \dfrac{1}{4} - \dfrac{3}{2} &= \dfrac{7}{8} + \dfrac{2}{4} - \dfrac{3}{2}\\
\\
&= \dfrac{7}{8} + \dfrac{1}{2} - \dfrac{3}{2}\\
\\
&= \dfrac{7}{8} + \dfrac{4}{8} - \dfrac{12}{8}\\
\\
&= \dfrac{7 + 4 - 12}{8}\\
\\
&= \dfrac{-1}{8}\\
\\
&= -\dfrac{1}{8}
\end{align*}
\end{split}\]
d)
\[
5 - \dfrac{2}{3} \cdot 4 + \dfrac{1}{6}
\]
\[
\dfrac{5}{2}
\]
\[\begin{split}
\begin{align*}
5 - \dfrac{2}{3} \cdot 4 + \dfrac{1}{6} &= 5 - \dfrac{8}{3} + \dfrac{1}{6}\\
\\
&= \dfrac{30}{6} - \dfrac{16}{6} + \dfrac{1}{6}\\
\\
&= \dfrac{30 - 16 + 1}{6}\\
\\
&= \dfrac{15}{6}\\
\\
&= \dfrac{5}{2}
\end{align*}
\end{split}\]
Oppgave 6#
Regn ut og skriv svaret så enkelt som mulig.
a)
\[
2 \cdot \dfrac{3}{4} + 5 \cdot \dfrac{1}{8}
\]
\[
\dfrac{17}{8}
\]
\[\begin{split}
\begin{align*}
2 \cdot \dfrac{3}{4} + 5 \cdot \dfrac{1}{8} &= \dfrac{6}{4} + \dfrac{5}{8}\\
\\
&= \dfrac{3}{2} + \dfrac{5}{8}\\
\\
&= \dfrac{12}{8} + \dfrac{5}{8}\\
\\
&= \dfrac{17}{8}
\end{align*}
\end{split}\]
b)
\[
\dfrac{1}{2} \cdot 3 + \dfrac{1}{4} \cdot 8
\]
\[
\dfrac{7}{2}
\]
\[\begin{split}
\begin{align*}
\dfrac{1}{2} \cdot 3 + \dfrac{1}{4} \cdot 8 &= \dfrac{3}{2} + \dfrac{8}{4}\\
\\
&= \dfrac{3}{2} + 2\\
\\
&= \dfrac{3}{2} + \dfrac{4}{2}\\
\\
&= \dfrac{7}{2}
\end{align*}
\end{split}\]
c)
\[
\dfrac{2}{3} \cdot \dfrac{3}{4} + \dfrac{1}{2} \cdot \dfrac{5}{6}
\]
\[
\dfrac{11}{12}
\]
\[\begin{split}
\begin{align*}
\dfrac{2}{3} \cdot \dfrac{3}{4} + \dfrac{1}{2} \cdot \dfrac{5}{6} &= \dfrac{6}{12} + \dfrac{5}{12}\\
\\
&= \dfrac{1}{2} + \dfrac{5}{12}\\
\\
&= \dfrac{6}{12} + \dfrac{5}{12}\\
\\
&= \dfrac{11}{12}
\end{align*}
\end{split}\]
d)
\[
\dfrac{1}{2} \cdot \dfrac{3}{4} - \dfrac{1}{3} \cdot \dfrac{2}{5}
\]
\[
\dfrac{29}{120}
\]
\[\begin{split}
\begin{align*}
\dfrac{1}{2} \cdot \dfrac{3}{4} - \dfrac{1}{3} \cdot \dfrac{2}{5} &= \dfrac{3}{8} - \dfrac{2}{15}\\
\\
&= \dfrac{3 \cdot 15}{8 \cdot 15} - \dfrac{2 \cdot 8}{15 \cdot 8}\\
\\
&= \dfrac{45}{120} - \dfrac{16}{120}\\
\\
&= \dfrac{29}{120}
\end{align*}
\end{split}\]
Oppgave 7#
Regn ut og forkort svaret så mye som mulig.
a)
\[
2 \cdot \left(\dfrac{3}{4} + \dfrac{4}{3}\right) : \dfrac{1}{4}
\]
\[
\dfrac{50}{3}
\]
\[\begin{split}
\begin{align*}
2 \cdot \left(\dfrac{3}{4} + \dfrac{4}{3}\right) : \dfrac{1}{4} &= 2 \cdot \left(\dfrac{9}{12} + \dfrac{16}{12}\right) : \dfrac{1}{4}\\
\\
&= 2 \cdot \dfrac{25}{12} : \dfrac{1}{4}\\
\\
&= \dfrac{50}{12} : \dfrac{1}{4}\\
\\
&= \dfrac{50}{12} \cdot \dfrac{4}{1}\\
\\
&= \dfrac{200}{12}\\
\\
&= \dfrac{50}{3}
\end{align*}
\end{split}\]
b)
\[
\dfrac{1}{2} \cdot \left(3 + 4\right) - 2 \cdot \dfrac{1}{4}
\]
\[
3
\]
\[\begin{split}
\begin{align*}
\dfrac{1}{2} \cdot \left(3 + 4\right) - 2 \cdot \dfrac{1}{4} &= \dfrac{1}{2} \cdot 7 - \dfrac{2}{4}\\
\\
&= \dfrac{7}{2} - \dfrac{1}{2}\\
\\
&= \dfrac{6}{2}\\
\\
&= 3
\end{align*}
\end{split}\]
c)
\[
\dfrac{2}{3} \cdot \left(\dfrac{3}{4} + \dfrac{1}{2}\right) + \dfrac{1}{5} : \dfrac{1}{3}
\]
\[
\dfrac{43}{30}
\]
\[\begin{split}
\begin{align*}
\dfrac{2}{3} \cdot \left(\dfrac{3}{4} + \dfrac{1}{2}\right) + \dfrac{1}{5} : \dfrac{1}{3} &= \dfrac{2}{3} \cdot \left(\dfrac{3}{4} + \dfrac{2}{4}\right) + \dfrac{1}{5} \cdot \dfrac{3}{1}\\
\\
&= \dfrac{2}{3} \cdot \dfrac{5}{4} + \dfrac{3}{5}\\
\\
&= \dfrac{10}{12} + \dfrac{3}{5}\\
\\
&= \dfrac{5}{6} + \dfrac{3}{5}\\
\\
&= \dfrac{25}{30} + \dfrac{18}{30}\\
\\
&= \dfrac{43}{30}
\end{align*}
\end{split}\]
d)
\[
\dfrac{1}{2} \cdot \left(\dfrac{3}{4} - \dfrac{1}{3}\right) + 2 : \dfrac{1}{5}
\]
\[
\dfrac{245}{24}
\]
\[\begin{split}
\begin{align*}
\dfrac{1}{2} \cdot \left(\dfrac{3}{4} - \dfrac{1}{3}\right) + 2 : \dfrac{1}{5} &= \dfrac{1}{2} \cdot \left(\dfrac{9}{12} - \dfrac{4}{12}\right) + 2 \cdot \dfrac{5}{1}\\
\\
&= \dfrac{1}{2} \cdot \dfrac{5}{12} + 10\\
\\
&= \dfrac{5}{24} + 10\\
\\
&= \dfrac{5}{24} + \dfrac{240}{24}\\
\\
&= \dfrac{245}{24}
\end{align*}
\end{split}\]
Oppgave 8#
Skriv kvadratrøttene så enkelt som mulig.
a)
\[
\sqrt{32}
\]
\[
4\sqrt{2}
\]
\[
\sqrt{32} = \sqrt{16 \cdot 2} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}
\]
b)
\[
\sqrt{50}
\]
\[
5\sqrt{2}
\]
\[
\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}
\]
c)
\[
\sqrt{27}
\]
\[
3\sqrt{3}
\]
\[
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
\]
d)
\[
\sqrt{18}
\]
\[
3\sqrt{2}
\]
\[
\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}
\]
Oppgave 9#
Skriv så enkelt som mulig.
a)
\[
\sqrt{32} + \sqrt{8}
\]
\[
6\sqrt{2}
\]
\[\begin{split}
\begin{align*}
\sqrt{32} + \sqrt{8} &= \sqrt{16 \cdot 2} + \sqrt{4 \cdot 2}\\
\\
&= 4\sqrt{2} + 2\sqrt{2}\\
\\
&= 6\sqrt{2}
\end{align*}
\end{split}\]
b)
\[
\sqrt{50} - \sqrt{2}
\]
\[
4\sqrt{2}
\]
\[\begin{split}
\begin{align*}
\sqrt{50} - \sqrt{2} &= \sqrt{25 \cdot 2} - \sqrt{2}\\
\\
&= 5\sqrt{2} - \sqrt{2}\\
\\
&= 4\sqrt{2}
\end{align*}
\end{split}\]
c)
\[
\sqrt{18} + \sqrt{72}
\]
\[
9\sqrt{2}
\]
\[\begin{split}
\begin{align*}
\sqrt{18} + \sqrt{72} &= \sqrt{9 \cdot 2} + \sqrt{36 \cdot 2}\\
\\
&= 3\sqrt{2} + 6\sqrt{2}\\
\\
&= 9\sqrt{2}
\end{align*}
\end{split}\]
d)
\[
\sqrt{27} - \sqrt{12}
\]
\[
\sqrt{3}
\]
\[\begin{split}
\begin{align*}
\sqrt{27} - \sqrt{12} &= \sqrt{9 \cdot 3} - \sqrt{4 \cdot 3}\\
\\
&= 3\sqrt{3} - 2\sqrt{3}\\
\\
&= \sqrt{3}
\end{align*}
\end{split}\]
Oppgave 10#
Skriv så enkelt som mulig.
a)
\[
\dfrac{\sqrt{32}}{\sqrt{8}}
\]
\[
2
\]
\[
\dfrac{\sqrt{32}}{\sqrt{8}} = \sqrt{\dfrac{32}{8}} = \sqrt{4} = 2
\]
b)
\[
\dfrac{\sqrt{50}}{\sqrt{2}}
\]
\[
5
\]
\[
\dfrac{\sqrt{50}}{\sqrt{2}} = \sqrt{\dfrac{50}{2}} = \sqrt{25} = 5
\]
c)
\[
\dfrac{\sqrt{18}}{\sqrt{72}}
\]
\[
\dfrac{1}{2}
\]
\[
\dfrac{\sqrt{18}}{\sqrt{72}} = \sqrt{\dfrac{18}{72}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}
\]
d)
\[
\dfrac{\sqrt{27}}{\sqrt{12}}
\]
\[
\dfrac{3}{2}
\]
\[
\dfrac{\sqrt{27}}{\sqrt{12}} = \sqrt{\dfrac{27}{12}} = \sqrt{\dfrac{9}{4}} = \dfrac{3}{2}
\]
Oppgave 11#
a)
\[
\sqrt{\dfrac{3}{4}}
\]
\[
\dfrac{\sqrt{3}}{2}
\]
\[
\sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{\sqrt{4}} = \dfrac{\sqrt{3}}{2}
\]
b)
\[
\sqrt{\dfrac{5}{9}}
\]
\[
\dfrac{\sqrt{5}}{3}
\]
\[
\sqrt{\dfrac{5}{9}} = \dfrac{\sqrt{5}}{\sqrt{9}} = \dfrac{\sqrt{5}}{3}
\]
c)
\[
\sqrt{\dfrac{8}{25}}
\]
\[
\dfrac{2\sqrt{2}}{5}
\]
\[
\sqrt{\dfrac{8}{25}} = \dfrac{\sqrt{8}}{\sqrt{25}} = \dfrac{\sqrt{4 \cdot 2}}{5} = \dfrac{2\sqrt{2}}{5}
\]
d)
\[
\sqrt{\dfrac{7}{16}}
\]
\[
\dfrac{\sqrt{7}}{4}
\]
\[
\sqrt{\dfrac{7}{16}} = \dfrac{\sqrt{7}}{\sqrt{16}} = \dfrac{\sqrt{7}}{4}
\]