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2.1 Exercises

From Förberedande kurs i matematik 1

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===Exercise 2.1:7===
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Current revision

       Theory          Exercises      


Exercise 2.1:1

Expand

a) 3x(x1) b) (1+xx2)xy c) x2(4y2)
d) x3y2y11xy+1  e) (x7)2 f) (5+4y)2
g) (y23x3)2 h) (5x3+3x5)2


Exercise 2.1:2

Expand

a) (x4)(x5)3x(2x3) b) (15x)(1+15x)3(25x)(2+5x)
c) (3x+4)2(3x2)(3x8) d) (3x2+2)(3x22)(9x4+4)
e) (a+b)2+(ab)2

Exercise 2.1:3

Factorise and simplify as much as possible

a) x236 b) 5x220 c) x2+6x+9
d) \displaystyle x^2-10x+25 e) \displaystyle 18x-2x^3 f) \displaystyle 16x^2+8x+1

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)}

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}

Exercise 2.1:7

Simplify the following by writing them as a single ordinary fraction

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 a single ordinary

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}}}