2.1 Exercises
From Förberedande kurs i matematik 1
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||<math> (5x^3+3x^5)^2</math> | ||<math> (5x^3+3x^5)^2</math> | ||
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| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:1|Solution a|Lösning 2.1:1a|Solution b|Lösning 2.1:1b|Solution c|Lösning 2.1:1c|Solution d|Lösning 2.1:1d|Solution e|Lösning 2.1:1e|Solution f|Lösning 2.1:1f|Solution g|Lösning 2.1:1g|Solution h|Lösning 2.1:1h}} |
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||<math> (a+b)^2+(a-b)^2</math> | ||<math> (a+b)^2+(a-b)^2</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:2|Solution a|Lösning 2.1:2a|Solution b|Lösning 2.1:2b|Solution c|Lösning 2.1:2c|Solution d|Lösning 2.1:2d|Solution e|Lösning 2.1:2e}} |
===Exercise 2.1:3=== | ===Exercise 2.1:3=== | ||
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||<math> 16x^2+8x+1</math> | ||<math> 16x^2+8x+1</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:3|Solution a|Lösning 2.1:3a|Solution b|Lösning 2.1:3b|Solution c|Lösning 2.1:3c|Solution d|Lösning 2.1:3d|Solution e|Lösning 2.1:3e|Solution f|Lösning 2.1:3f}} |
===Exercise 2.1:4=== | ===Exercise 2.1:4=== | ||
<div class="ovning"> | <div class="ovning"> | ||
| - | Determine the coefficients in front of <math>\,x\,</math> and <math>\,x^2\</math> when the following | + | Determine the coefficients in front of <math>\,x\,</math> and <math>\,x^2\</math> when the following expressions are expanded out. |
{| width="100%" cellspacing="10px" | {| width="100%" cellspacing="10px" | ||
|a) | |a) | ||
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|- | |- | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:4|Solution a|Lösning 2.1:4a|Solution b|Lösning 2.1:4b|Solution c|Lösning 2.1:4c}} |
===Exercise 2.1:5=== | ===Exercise 2.1:5=== | ||
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|| <math>\displaystyle \frac{(y^2+4y+4)(2y-4)}{(y^2+4)(y^2-4)}</math> | || <math>\displaystyle \frac{(y^2+4y+4)(2y-4)}{(y^2+4)(y^2-4)}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:5|Solution a|Lösning 2.1:5a|Solution b|Lösning 2.1:5b|Solution c|Lösning 2.1:5c|Solution d|Lösning 2.1:5d}} |
===Exercise 2.1:6=== | ===Exercise 2.1:6=== | ||
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|| <math>\displaystyle\frac{a-b+\displaystyle\frac{b^2}{a+b}}{1-\left(\displaystyle\frac{a-b}{a+b}\right)^2}</math> | || <math>\displaystyle\frac{a-b+\displaystyle\frac{b^2}{a+b}}{1-\left(\displaystyle\frac{a-b}{a+b}\right)^2}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:6|Solution a|Lösning 2.1:6a|Solution b|Lösning 2.1:6b|Solution c|Lösning 2.1:6c|Solution d|Lösning 2.1:6d}} |
===Exercise 2.1:7=== | ===Exercise 2.1:7=== | ||
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|width="33%" | <math>\displaystyle \frac{ax}{a+1}-\displaystyle \frac{ax^2}{(a+1)^2}</math> | |width="33%" | <math>\displaystyle \frac{ax}{a+1}-\displaystyle \frac{ax^2}{(a+1)^2}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:7|Solution a|Lösning 2.1:7a|Solution b|Lösning 2.1:7b|Solution c|Lösning 2.1:7c}} |
===Exercise 2.1:8=== | ===Exercise 2.1:8=== | ||
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|width="33%" | <math>\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+x}}}</math> | |width="33%" | <math>\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+x}}}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Answer|Svar 2.1:8|Solution a|Lösning 2.1:8a|Solution b|Lösning 2.1:8b|Solution c|Lösning 2.1:8c}} |
Revision as of 12:49, 18 August 2008
| Theory | Exercises |
Exercise 2.1:1
Expand
| a) | \displaystyle 3x(x-1) | b) | \displaystyle (1+x-x^2)xy | c) | \displaystyle -x^2(4-y^2) |
| d) | \displaystyle x^3y^2\left(\displaystyle \frac{1}{y} - \frac{1}{xy}+1\right) | e) | \displaystyle (x-7)^2 | f) | \displaystyle (5+4y)^2 |
| g) | \displaystyle (y^2-3x^3)^2 | h) | \displaystyle (5x^3+3x^5)^2 |
Answer
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Solution a
Solution b
Solution c
Solution d
Solution e
Solution f
Solution g
Solution h
Exercise 2.1:2
Expand
| a) | \displaystyle (x-4)(x-5)-3x(2x-3) | b) | \displaystyle (1-5x)(1+15x)-3(2-5x)(2+5x) |
| c) | \displaystyle (3x+4)^2-(3x-2)(3x-8) | d) | \displaystyle (3x^2+2)(3x^2-2)(9x^4+4) |
| e) | \displaystyle (a+b)^2+(a-b)^2 |
Answer
Solution a
Solution b
Solution c
Solution d
Solution e
Exercise 2.1:3
Factorise and simplify as much as possible
| a) | \displaystyle x^2-36 | b) | \displaystyle 5x^2-20 | c) | \displaystyle x^2+6x+9 |
| d) | \displaystyle x^2-10x+25 | e) | \displaystyle 18x-2x^3 | f) | \displaystyle 16x^2+8x+1 |
Answer
Solution a
Solution b
Solution c
Solution d
Solution e
Solution f
Exercise 2.1:4
Determine the coefficients in front of \displaystyle \,x\, and \displaystyle \,x^2\ when the following expressions are expanded out.
| a) | \displaystyle (x+2)(3x^2-x+5) |
| b) | \displaystyle (1+x+x^2+x^3)(2-x+x^2+x^4) |
| c) | \displaystyle (x-x^3+x^5)(1+3x+5x^2)(2-7x^2-x^4) |
Exercise 2.1:5
Simplify as much as possible
| a) | \displaystyle \displaystyle \frac{1}{x-x^2}-\displaystyle \frac{1}{x} | b) | \displaystyle \displaystyle \frac{1}{y^2-2y}-\displaystyle \frac{2}{y^2-4} |
| c) | \displaystyle \displaystyle \frac{(3x^2-12)(x^2-1)}{(x+1)(x+2)} | d) | \displaystyle \displaystyle \frac{(y^2+4y+4)(2y-4)}{(y^2+4)(y^2-4)} |
Answer
Solution a
Solution b
Solution c
Solution d
Exercise 2.1:6
Simplify as much as possible
| a) | \displaystyle \left(x-y+\displaystyle\frac{x^2}{y-x}\right) \displaystyle \left(\displaystyle\frac{y}{2x-y}-1\right) | b) | \displaystyle \displaystyle \frac{x}{x-2}+\displaystyle \frac{x}{x+3}-2 |
| c) | \displaystyle \displaystyle \frac{2a+b}{a^2-ab}-\frac{2}{a-b} | d) | \displaystyle \displaystyle\frac{a-b+\displaystyle\frac{b^2}{a+b}}{1-\left(\displaystyle\frac{a-b}{a+b}\right)^2} |
Answer
Solution a
Solution b
Solution c
Solution d
Exercise 2.1:7
Simplify the following fractions by writing them as an expression having a common fraction sign
| a) | \displaystyle \displaystyle \frac{2}{x+3}-\frac{2}{x+5} | b) | \displaystyle x+\displaystyle \frac{1}{x-1}+\displaystyle \frac{1}{x^2} | c) | \displaystyle \displaystyle \frac{ax}{a+1}-\displaystyle \frac{ax^2}{(a+1)^2} |
Exercise 2.1:8
Simplify the following fractions by writing them as an expression having a common fraction sign
| a) | \displaystyle \displaystyle \frac{\displaystyle\ \frac{x}{x+1}\ }{\ 3+x\ } | b) | \displaystyle \displaystyle \frac{\displaystyle \frac{3}{x}-\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x-3}} | c) | \displaystyle \displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+x}}} |
