Facit

Sommarmatte 1

(Skillnad mellan versioner)

Versionen från 16 juli 2007 kl. 08.36

Svar 1.1:1

a)	-7	b)	1
c)	11	d)	1

Svar 1.1:2

a)	0	b)	-1
c)	-25	d)	-19

Svar 1.1:3


a)	naturliga talen, heltalen, rationella talen	b)	heltalen, rationella talen	c)	naturliga talen, heltalen, rationella talen
d)	heltalen, rationella talen	e)	heltalen, rationella talen	f)	naturliga talen, heltalen, rationella talen
g)	rationella talen	h)	naturliga talen, heltalen, rationella talen	i)	irrationella talen
j)	naturliga talen, heltalen, rationella talen	k)	irrationella talen	l)	irrationella talen

Svar 1.1:4

a)	\displaystyle \frac{3}{5}<\frac{5}{3}<2<\frac{7}{3}
b)	\displaystyle -\frac{1}{2}<-\frac{1}{3}<-\frac{3}{10}<-\frac{1}{5}
c)	\displaystyle \frac{1}{2}<\frac{3}{5}<\frac{21}{34}<\frac{5}{8}<\frac{2}{3}

Svar 1.1:5

a)	1{,}167	b)	2{,}250	c)	0{,}286	d)	1{,}414

Svar 1.1:6

a)	Talet är rationellt och lika med \,314/100 = 157/50\,.
b)	Talet är rationellt och är lika med \,31413/9999 = 10471/3333\,.
c)	Talet är rationellt och lika med \,1999/9990\,.
d)	Talet är irrationellt.

Svar 1.2:1

a)	\displaystyle \frac{93}{28}	b)	\displaystyle \frac{3}{35}	c)	\displaystyle -\frac{7}{30}
d)	\displaystyle \frac{47}{60}	e)	\displaystyle \frac{47}{84}

Svar 1.2:2

a)	\displaystyle {30}	b)	\displaystyle {8}
c)	\displaystyle {84}	d)	\displaystyle {225}

Svar 1.2:3


a)	\displaystyle \frac{19}{100}	b)	\displaystyle \frac{1}{240}

Svar 1.2:4

a)	\displaystyle \frac{6}{7}	b)	\displaystyle \frac{16}{21}	c)	\displaystyle \frac{1}{6}

Övning 1.2:5


a)	\displaystyle \frac{105}{4}	b)	-5	c)	\displaystyle \frac{8}{55}

Svar 1.2:6

\displaystyle \frac{152}{35}

Övning 1.3:1

a)	72	b)	3	c)	-125	d)	\displaystyle \frac{27}{8}

Övning 1.3:2

a)	2^6	b)	2^{-2}	c)	2^0

Övning 1.3:3

a)	3^{-1}	b)	3^5	c)	3^4	d)	3^{-3}	e)	3^{-3}

Övning 1.3:4

a)	4	b)	3	c)	625
d)	16	e)	\displaystyle \frac{1}{3750}

Övning 1.3:5

a)	2	b)	\displaystyle \frac{1}{2}	c)	27
d)	2209	e)	9	f)	\displaystyle \frac{25}{3}

Övning 1.3:6

a)	256^{1/3}>200^{1/3}	b)	0{,}4^{-3}>0{,}5^{-3}	c)	0{,}2^{5}>0{,}2^{7}
d)	\bigl(5^{1/3}\bigr)^{4}>400^{1/3}	e)	125^{1/2}>625^{1/3}	f)	3^{40}>2^{56}

Svar 2.1:1

a)	3x^2-3x	b)	xy+x^2y-x^3y	c)	-4x^2+x^2y^2
d)	x^3y-x^2y+x^3y^2	e)	x^2-14x+49	f)	16y^2+40y+25
g)	9x^6-6x^3y^2+y^4	h)	9x^{10}+30x^8+25x^6

Svar 2.1:2

a)	-5x^2+20	b)	10x-11
c)	54x	d)	81x^8-16
e)	2a^2+2b^2

Svar 2.1:3

a)	(x+6)(x-6)	b)	5(x+2)(x-2)	c)	(x+3)^2
d)	(x-5)^2	e)	-2x(x+3)(x-3)	f)	(4x+1)^2

Svar 2.1:4

a)	5\, framför \,x^2\,, \,3\, framför \,x
b)	2\, framför \,x^2\,, \,1\, framför \,x
\textrm{c) }	6\, framför \,x^2\,, \,2\, framför \,x

Svar 2.1:5

a)	\displaystyle \frac{1}{1-x}	b)	-\displaystyle \frac{1}{y(y+2)}
c)	3(x-2)(x-1)	d)	\displaystyle \frac{2(y+2)}{y^2+4}

Svar 2.1:6

a)	2y	b)	\displaystyle\frac{-x+12}{(x-2)(x+3)}
c)	\displaystyle\frac{b}{a(a-b)}	d)	\displaystyle\frac{a(a+b)}{4b}

Svar 2.1:7

\displaystyle \frac{4}{(x+3)(x+5)}

\displaystyle \frac{x^4-x^3+x^2+x-1}{x^2(x-1)}

\displaystyle \frac{ax(a+1-x)}{(a+1)^2}

Svar 2.1:8

a)	\displaystyle \frac{x}{(x+3)(x+1)}	b)	\displaystyle \frac{2(x-3)}{x}	c)	\displaystyle \frac{x+2}{2x+3}

Svar 2.2:1

a)	x=1	b)	x=6
c)	x=-\displaystyle\frac{3}{2}	d)	x=-\displaystyle\frac{13}{3}

Svar 2.2:2

a)	x=1	b)	x=\displaystyle\frac{5}{3}
c)	x=2	d)	x=-2

Svar 2.2:3

a)	x=9
b)	x=\displaystyle\frac{7}{5}
c)	x=\displaystyle\frac{4}{5}
d)	x=\displaystyle\frac{1}{2}

Övning 2.2:4

a)	-2x+y=3
b)	y=-\displaystyle\frac{3}{4}x+\frac{5}{4}

Övning 2.2:5

a)	y=-3x+9
b)	y=-3x+1
c)	y=3x+5
d)	y=-\displaystyle \frac{1}{2}x+5
e)	k = \displaystyle\frac{8}{5}

Övning 2.2:6

a)	\bigl(-\frac{5}{3},0\bigr)	b)	(0,5)
c)	\bigl(0,-\frac{6}{5}\bigr)	d)	(12,-13)
e)	\bigl(-\frac{1}{4},\frac{3}{2}\bigr)

Övning 2.2:7

a)		b)
c)

Övning 2.2:8

a)		b)
c)

Övning 2.2:9

a)	4\, a.e.
b)	5\, a.e.
c)	6\, a.e.

Övning 2.3:1

a)	(x-1)^2-1	b)	(x+1)^2-2	c)	-(x-1)^2+6	d)	\bigl(x+\frac{5}{2}\bigr)^2-\frac{13}{4}

Övning 2.3:2


a)	\left\{ \eqalign{ x_1 &= 1 \cr x_2 &= 3\cr }\right.	b)	\left\{ \eqalign{ y_1 &= -5 \cr y_2 &= 3\cr }\right.	c)	saknar (reella) lösning
d)	\left\{ \eqalign{ x_1 &= \textstyle\frac{1}{2}\cr x_2 &= \textstyle\frac{13}{2}\cr }\right.	e)	\left\{ \eqalign{ x_1 &= -1 \cr x_2 &= \textstyle\frac{3}{5}\cr }\right.	f)	\left\{ \eqalign{ x_1 &= \textstyle\frac{4}{3}\cr x_2 &= 2\cr }\right.

Övning 2.3:3

a)	\left\{ \eqalign{ x_1 &= 0 \cr x_2 & = -3\cr }\right.	b)	\left\{ \eqalign{ x_1 &= 3 \cr x_2 & = -5\cr }\right.
c)	\left\{ \eqalign{ x_1 & = \textstyle\frac{2}{3} \cr x_2 & = -8\cr }\right.	d)	\left\{ \eqalign{ x_1 & = 0\cr x_2 & = 12\cr }\right.
e)	\left\{ \eqalign{ x_1 & = -3 \cr x_2 & = 8\cr }\right.	f)	\left\{ \eqalign{ x_1 & = 0 \cr x_2 & = 1 \cr x_3 & = 2 }\right.

Övning 2.3:4

a)	ax^2-ax-2a=0\,, där \,a\ne 0\, är en konstant.
b)	ax^2-2ax-2a=0\,, där \,a\ne 0\, är en konstant.
c)	ax^2-(3+\sqrt{3}\,)ax+3\sqrt{3}\,a=0\,, där \,a\ne 0\, är en konstant.

Övning 2.3:5

a)	Exempelvis \ x^2+14x+49=0\,.
b)	3< x<4
c)	b=-5

Övning 2.3:6

a)	0	b)	-2	c)	\displaystyle \frac{3}{4}

Övning 2.3:7

a)	1	b)	\displaystyle -\frac{7}{4}	c)	saknar max

Övning 2.3:8

Se lösningen i webmaterialet när du loggat in till kursen.

Övning 2.3:9

a)	(-1,0)\ och \ (1,0)	b)	(2,0)\ och \ (3,0)	c)	(1,0)\ och \ (3,0)

Övning 2.3:10

Se lösningen i webmaterialet när du loggat in till kursen

Svar 3.1:1

a)	2^{1/2}	b)	7^{5/2}	c)	3^{4/3}	d)	3^{1/4}

Svar 3.1:2

a)	3	b)	3	c)	ej definierad	d)	5^{11/6}
e)	12	f)	2	g)	-5

Svar 3.1:3

a)	3	b)	\displaystyle \frac{4\sqrt{3}}{3}
c)	2\sqrt{5}	d)	2-\sqrt{2}

Svar 3.1:4

a)	0{,}4	b)	0{,}3
c)	-4\sqrt{2}	d)	2\sqrt{3}

Svar 3.1:5

a)	\displaystyle \frac{\sqrt{3}}{3}	b)	\displaystyle \frac{7^{2/3}}{7}	c)	3-\sqrt{7}	d)	\displaystyle \frac{\sqrt{17}+\sqrt{13}}{4}

Svar 3.1:6

a)	6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}	b)	-\displaystyle \frac{5+4\sqrt{3}}{23}
c)	\displaystyle \frac{2}{3}\sqrt{6}+\displaystyle \frac{2}{3}\sqrt{3}-\displaystyle \frac{2}{5}\sqrt{10}-\displaystyle \frac{2}{5}\sqrt{5}	d)	\displaystyle \frac{5\sqrt{3}+7\sqrt{2}-\sqrt{6}-12}{23}

Svar 3.1:7

a)	\sqrt{5}-\sqrt{7}	b)	-\sqrt{35}	c)	\sqrt{17}

Svar 3.1:8

a)	\sqrt[\scriptstyle3]6 > \sqrt[\scriptstyle3]5	b)	7 > \sqrt7
c)	\sqrt7 > 2{,}5	d)	\sqrt[\scriptstyle3]2\cdot3 > \sqrt2\bigl(\sqrt[\scriptstyle4]3\,\bigr)^3

Svar 3.2:1

x=5

Svar 3.2:2

x=1

Svar 3.2:3

\left \{ \eqalign{ x_1 & = 3 \cr x_2 & = 4\cr } \right.

Svar 3.2:4

Saknar lösning.

Svar 3.2:5

x=1

Svar 3.2:6

x=\displaystyle\frac{5}{4}

Svar 3.3:1

a)	x=3	b)	x=-1
c)	x=-2	d)	x=4

Svar 3.3:2

a)	-1	b)	4	c)	-3	d)	0
e)	2	f)	3	g)	10	h)	-2

Svar 3.3:3

a)	3	b)	-\displaystyle \frac{1}{2}	c)	-3
d)	\displaystyle \frac{7}{3}	e)	4	f)	-2
g)	1	h)	\displaystyle \frac{5}{2}

Svar 3.3:4

a)	1	b)	0	c)	-\displaystyle \frac{1}{2}\lg{3}

Svar 3.3:5

a)	5	b)	0	c)	0
d)	0	e)	-2	f)	e^2

Svar 3.3:6


a)	1{,}262
b)	1{,}663
c)	4{,}762

Svar 3.4:1

a)	x=\ln 13	b)	x=\displaystyle\frac{\ln 2 - \ln 13}{1+\ln 3}	c)	x=\displaystyle\frac{\ln 7 - \ln 3}{1-\ln 2}

Svar 3.4:2

a)	\left\{ \eqalign{ x_1&=\sqrt2 \cr x_2&=-\sqrt2 } \right.	b)	x=\ln \left(\displaystyle\frac{\sqrt17}{2}-\frac{1}{2}\right)	c)	Saknar lösning

Svar 3.4:3

a)	x=-\,\displaystyle\frac{1}{\ln{2}}\pm\sqrt{\left(\displaystyle\frac{1}{\ln{2}}\right)^2-1}	b)	x=\displaystyle \frac{5}{2}
c)	x=1

Den här artikeln är hämtad från http://wiki.math.se/wikis/sf0600_0701/index.php/Facit

 <td>Se lösningen i webmaterialet när du loggat in till kursen</td>
 </tr>
+</table>
+</div>
+==Svar 3.1:1==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="25%">2^{1/2}</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="25%">7^{5/2}</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="25%">3^{4/3}</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="25%">3^{1/4}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.1:2==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="25%">3</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="25%">3</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="25%">ej definierad</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="25%">5^{11/6}</td>
+</tr>
+<tr align="left">
+<td class="ntext">e)</td>
+<td class="ntext" width="25%">12</td>
+<td class="ntext">f)</td>
+<td class="ntext" width="25%">2</td>
+<td class="ntext">g)</td>
+<td class="ntext" width="25%">-5</td>
+<td></td>
+<td></td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.1:3==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="50%">3</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="50%">\displaystyle \frac{4\sqrt{3}}{3}</td>
+</tr>
+<tr align="left">
+<td class="ntext">c)</td>
+<td class="ntext" width="50%">2\sqrt{5}</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="50%">2-\sqrt{2}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.1:4==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="50%">0{,}4</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="50%">0{,}3</td>
+</tr>
+<tr align="left">
+<td class="ntext">c)</td>
+<td class="ntext" width="50%">-4\sqrt{2}</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="50%">2\sqrt{3}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.1:5==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="25%">\displaystyle \frac{\sqrt{3}}{3}</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="25%">\displaystyle \frac{7^{2/3}}{7}</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="25%">3-\sqrt{7}</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="25%">\displaystyle \frac{\sqrt{17}+\sqrt{13}}{4}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.1:6==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="50%">6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="50%">-\displaystyle \frac{5+4\sqrt{3}}{23}</td>
+</tr>
+<tr align="left">
+<td class="ntext">c)</td>
+<td class="ntext" width="50%">\displaystyle \frac{2}{3}\sqrt{6}+\displaystyle \frac{2}{3}\sqrt{3}-\displaystyle \frac{2}{5}\sqrt{10}-\displaystyle \frac{2}{5}\sqrt{5}</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="50%">\displaystyle \frac{5\sqrt{3}+7\sqrt{2}-\sqrt{6}-12}{23}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.1:7==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="33%">\sqrt{5}-\sqrt{7}</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="33%">-\sqrt{35}</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="33%">\sqrt{17}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.1:8==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="50%">\sqrt[\scriptstyle3]6 > \sqrt[\scriptstyle3]5</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="50%">7 > \sqrt7</td>
+</tr>
+<tr align="left">
+<td class="ntext">c)</td>
+<td class="ntext" width="50%">\sqrt7 > 2{,}5</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="50%">\sqrt[\scriptstyle3]2\cdot3 > \sqrt2\bigl(\sqrt[\scriptstyle4]3\,\bigr)^3</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.2:1==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr><td height="5px"/></tr>
+<tr align="left">
+<td class="ntext" width="100%">x=5</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.2:2==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr><td height="5px"/></tr>
+<tr align="left">
+<td class="ntext" width="100%">x=1</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.2:3==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr><td height="5px"/></tr>
+<tr align="left">
+<td class="ntext" width="100%">\left \{ \eqalign{ x_1 & = 3 \cr x_2 & = 4\cr } \right.</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.2:4==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr><td height="5px"/></tr>
+<tr align="left">
+<td class="ntext" width="100%">Saknar l&ouml;sning.</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.2:5==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr><td height="5px"/></tr>
+<tr align="left">
+<td class="ntext" width="100%">x=1</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.2:6==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr><td height="5px"/></tr>
+<tr align="left">
+<td class="ntext" width="100%">x=\displaystyle\frac{5}{4}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.3:1==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="50%">x=3</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="50%">x=-1</td>
+</tr>
+<tr align="left">
+<td class="ntext">c)</td>
+<td class="ntext" width="50%">x=-2</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="50%">x=4</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.3:2==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="25%">-1</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="25%">4</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="25%">-3</td>
+<td class="ntext">d)</td>
+<td class="ntext" width="25%">0</td>
+</tr>
+<tr align="left">
+<td class="ntext">e)</td>
+<td class="ntext" width="25%">2</td>
+<td class="ntext">f)</td>
+<td class="ntext" width="25%">3</td>
+<td class="ntext">g)</td>
+<td class="ntext" width="25%">10</td>
+<td class="ntext">h)</td>
+<td class="ntext" width="25%">-2</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.3:3==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="33%">3</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="33%">-\displaystyle \frac{1}{2}</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="33%">-3</td>
+</tr>
+<tr align="left">
+<td class="ntext">d)</td>
+<td class="ntext" width="33%">\displaystyle \frac{7}{3}</td>
+<td class="ntext">e)</td>
+<td class="ntext" width="33%">4</td>
+<td class="ntext">f)</td>
+<td class="ntext" width="33%">-2</td>
+</tr>
+<tr align="left">
+<td class="ntext">g)</td>
+<td class="ntext" width="33%">1</td>
+<td class="ntext">h)</td>
+<td class="ntext" width="33%">\displaystyle \frac{5}{2}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.3:4==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="33%">1</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="33%">0</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="33%">-\displaystyle \frac{1}{2}\lg{3}</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.3:5==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="33%">5</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="33%">0</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="33%">0</td>
+</tr>
+<tr align="left">
+<td class="ntext">d)</td>
+<td class="ntext" width="33%">0</td>
+<td class="ntext">e)</td>
+<td class="ntext" width="33%">-2</td>
+<td class="ntext">f)</td>
+<td class="ntext" width="33%">e^2</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.3:6==
+<div class="ovning">
+<table width="100%"  cellspacing="10px">
+<tr align="left"><td height="5px"/></tr>
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="100%">1{,}262</td>
+</tr>
+<tr align="left">
+<td class="ntext">b)</td>
+<td class="ntext" width="100%">1{,}663</td>
+</tr>
+<tr align="left">
+<td class="ntext">c)</td>
+<td class="ntext" width="100%">4{,}762</td>
+</tr>
+<tr align="left"><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.4:1==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="33%">x=\ln 13</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="33%">x=\displaystyle\frac{\ln 2 - \ln 13}{1+\ln 3}</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="33%">x=\displaystyle\frac{\ln 7 - \ln 3}{1-\ln 2}</td>
+<tr><td height="5px"/></tr>
+</tr>
+</table>
+</div>
+==Svar 3.4:2==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="33%"> \left\{ \eqalign{ x_1&=\sqrt2 \cr x_2&=-\sqrt2 } \right. </td>
+<td class="ntext">b)</td>
+<td class="ntext" width="33%">x=\ln \left(\displaystyle\frac{\sqrt17}{2}-\frac{1}{2}\right)</td>
+<td class="ntext">c)</td>
+<td class="ntext" width="33%">Saknar l&ouml;sning</td>
+</tr>
+<tr><td height="5px"/></tr>
+</table>
+</div>
+==Svar 3.4:3==
+<div class="ovning">
+<table width="100%" cellspacing="10px">
+<tr align="left">
+<td class="ntext">a)</td>
+<td class="ntext" width="50%">x=-\,\displaystyle\frac{1}{\ln{2}}\pm\sqrt{\left(\displaystyle\frac{1}{\ln{2}}\right)^2-1}</td>
+<td class="ntext">b)</td>
+<td class="ntext" width="50%">x=\displaystyle \frac{5}{2}</td>
+</tr>
+<tr align="left">
+<td class="ntext">c)</td>
+<td class="ntext" width="50%">x=1</td>
+</tr>
+<tr><td height="5px"/></tr>
 </table>
 </div>