5. Exercises
From Förberedande kurs i matematik 1
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| - | __NOTOC__ | ||
| - | {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | ||
| - | | style="border-bottom:1px solid #000" width="5px" | | ||
| - | {{Not selected tab|[[5. Written presentation and communication of mathematics|Theory]]}} | ||
| - | {{Selected tab|[[5. Exercises|Exercises]]}} | ||
| - | | style="border-bottom:1px solid #000" width="100%"| | ||
| - | |} | ||
| - | |||
| - | ===Exercise 5:1=== | ||
| - | <div class="ovning"> | ||
| - | Write the formulas in TeX , you can test your text in the editor to your individual assignment. | ||
| - | {| width="100%" cellspacing="10px" | ||
| - | |a) | ||
| - | |width="50%" | <math>\displaystyle\frac{a+b}{d-c}</math> | ||
| - | |b) | ||
| - | |width="50%" | <math>\sqrt{3}</math> | ||
| - | |- | ||
| - | |c) | ||
| - | || <math>4x^2 - x</math> | ||
| - | |d) | ||
| - | || <math>\sin^2 x + \cos x</math> | ||
| - | |} | ||
| - | </div>{{#NAVCONTENT:Answer|Answer 5.1:1}} | ||
| - | |||
| - | |||
| - | ===Exercise 5:2=== | ||
| - | <div class="ovning"> | ||
| - | Write the formulas in TeX | ||
| - | {| width="100%" cellspacing="10px" | ||
| - | |a) | ||
| - | |width="50%" | <math>\cos{v} = \cos{\displaystyle \frac{3\pi}{2}}</math> | ||
| - | |b) | ||
| - | |width="50%" | <math>\tan\displaystyle\frac{u}{2}=\frac{\sin u}{1+\cos u}</math> | ||
| - | |- | ||
| - | |c) | ||
| - | || <math>\left\{\eqalign{ | ||
| - | x&=n\pi\cr | ||
| - | x&=\displaystyle \frac{\pi}{4}+\displaystyle \frac{n\pi}{2} | ||
| - | }\right.</math> | ||
| - | |d) | ||
| - | || <math>\bigl(\sqrt[\scriptstyle4]3\,\bigr)^3\ \sqrt{2 + \sqrt{4}}</math> | ||
| - | |} | ||
| - | </div>{{#NAVCONTENT:Answer|Answer 5.1:2}} | ||
| - | |||
| - | ===Exercise 5:3=== | ||
| - | <div class="ovning"> | ||
| - | Write the formulas in TeX | ||
| - | {| width="100%" cellspacing="10px" | ||
| - | |a) | ||
| - | |width="50%" | <math>x^3y^2\left(\displaystyle \frac{1}{y} - \frac{1}{xy}+1\right)</math> | ||
| - | |b) | ||
| - | |width="50%" | <math>\left(x-y+\displaystyle\frac{x^2}{y-x}\right) \left(\displaystyle\frac{y}{2x-y}-1\right)</math> | ||
| - | |- | ||
| - | |c) | ||
| - | || <math>\displaystyle\frac{a-b+\displaystyle\frac{b^2}{a+b}}{1-\left(\displaystyle\frac{a-b}{a+b}\right)^2}</math> | ||
| - | |d) | ||
| - | || <math>\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+x}}}</math> | ||
| - | |} | ||
| - | </div>{{#NAVCONTENT:Answer|Answer 5.1:3}} | ||
