Answer 5.1:3

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
(New page: {| 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+\displ...)
Current revision (14:20, 28 January 2009) (edit) (undo)
 
Line 1: Line 1:
{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
-
|width="50%" | <math>x^3y^2\left(\displaystyle \frac{1}{y} - \frac{1}{xy}+1\right)</math>
+
|width="100%" | \dfrac{x+1}{x^2-1} = \dfrac{1}{x-1}
 +
|-
|b)
|b)
-
|width="50%" | <math>\left(x-y+\displaystyle\frac{x^2}{y-x}\right) \left(\displaystyle\frac{y}{2x-y}-1\right)</math>
+
|width="100%" | \left(\dfrac{5}{x}-1\right)(1-x)
|-
|-
|c)
|c)
-
|| <math>\displaystyle\frac{a-b+\displaystyle\frac{b^2}{a+b}}{1-\left(\displaystyle\frac{a-b}{a+b}\right)^2}</math>
+
|width="100%" | \dfrac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}
 +
|-
|d)
|d)
-
|| <math>\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+x}}}</math>
+
|width="100%" | \dfrac{1}{1+\dfrac{1}{1+x}}
|}
|}

Current revision

a) \dfrac{x+1}{x^2-1} = \dfrac{1}{x-1}
b) \left(\dfrac{5}{x}-1\right)(1-x)
c) \dfrac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}
d) \dfrac{1}{1+\dfrac{1}{1+x}}
Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Answer_5.1:3"