Question

attached file

Answer 1

change it to multiplication and flip the second fraction:\[\frac{x^2+9x+20}{x^2-25} \div \frac{x+4}{x-4} = \frac{x^2+9x+20}{x^2-25} \times \frac{x-4}{x+4}\]
factorise:\[\frac{x^2+9x+20}{x^2-25} \times \frac{x-4}{x+4} = \frac{(x+4)(x+5)}{(x+5)(x-5)} \times \frac{x-4}{x+4}\] now multiply across top and bottom (as usual for fractions)\[\frac{(x+4)(x+5)(x-4)}{(x+5)(x-5)(x+4)} \]now cancel things which are on the top and bottom:
\[\frac{x-4}{x-5}\] and if you want:
\[1 + \frac{1}{x-5}\]

Answer 2

the important trick here is "difference of two squares" notice:
\[(x+y)(x-y) = x^2 - y^2\]

with part 2, rewrite it out so it looks like how Part 1 started, so we dont have horrible fractions-on-fractions

Answer 3

or.. multiply the top and bottom of the big fraction by \[(x-2)(x^2 - 4)\] this will also get rid of that nasty fraction tower..

Answer 4

this is problem with the 2x

Answer 5

what do you mean?

Answer 6

is this the x2+2x+1 problem

Answer 7

ah well can you see how to factorise x^2 + 2x +1 ?

Answer 8

its going to go into form (Ax + B)(Cx + D)

can you see that A and C have to be 1 to make the x^2 ?

Answer 9

no which problem where u talken about

Answer 10

\[\frac{x^2 + 2x +1}{x-2} \div \frac{x^2 -1}{x^2-4}\]


\[\frac{(x+1)(x+1)}{x-2} \div \frac{x^2 -1}{x^2-4}\]

Answer 11

\[\frac{(x+1)(x+1)}{x-2} \div \frac{(x+1)(x-1)}{x^2-4}\]

Answer 12

\[\frac{(x+1)(x+1)}{x-2} \div \frac{(x+1)(x-1)}{(x+2)(x-2)}\]

Answer 13

\[\frac{(x+1)(x+1)}{x-2} \times \frac{(x+2)(x-2)}{(x+1)(x-1)}\]

Answer 14

i think you just need to practice the factorising bit