Question

Simplify the complex fraction:

(Problem in comments...)

Answer 1

\[\frac{ \frac{ \frac{ 1 }{ a^2+11a+18 } }{ 1 } }{ a+2 }\]

Answer 2

start by factorising \[a^2+11a+18=(a+?)(a+??)\]

Answer 3

9 & 2

Answer 4

good.
\[a^2+11a+18=(a+9)(a+2)\]

Answer 5

What's the next step?

Answer 6

Now lets look at the top of the big fraction\[\frac{ \frac{ 1 }{ (a+9)(a+2)} }{ 1 }\]

how can we simplify a number divided by 1?

Answer 7

it cancels out and you just have the number on top?

Answer 8

right so the top of the big fraction
simplifies to become

\[\frac{ 1 }{ (a+9)(a+2)}\]

Answer 9

Now you have \[\frac{\frac{ 1 }{ (a+9)(a+2)} }{a+2}\]

Answer 10

the a+2 cancels out on the top & bottom?

Answer 11

divide the numerator and the denominator of this big fraction by a+2

(it doesn't cancel the a+2 in the numerator, because the a+2 is in the denominator of the little fraction in the numerator of the big fraction )

Answer 12

I'll just be left with 1/a+9?

Answer 13

nope it doesn't cancel

\[\frac{\frac{ 1 }{ (a+9)(a+2)} }{a+2}=\frac{\frac{ 1 }{ (a+9)(a+2)}\div(a+2) }{(a+2)\div(a+2)}=\]

Answer 14

\[\qquad\qquad\qquad\qquad=\frac{\frac{ 1 }{ (a+9)(a+2)}\times\frac1{a+2} }{(a+2)\times\frac1{a+2}}=\]

Answer 15

simplify the denominator first,
what does \[(a+2)\times\frac1{a+2}\]
simplify to become?

Answer 16

(this bit does cancel )

Answer 17

@cookiibabii93

Answer 18

I'm back...

Answer 19

can you simplify the denominator of the big fraction
\((a+2)\times\frac1{a+2} = \)

Answer 20

how do I simplify it?

Answer 21

\[(a+2)\times\frac1{a+2} =\frac{a+2}{a+2}=\frac{\cancel{a+2}}{\cancel{a+2}}=1\]

Answer 22

I thought so, but was unsure, lol thank you so much :)

Answer 23

so your left with \[\frac{ 1 }{ (a+9)(a+2)}\times\frac1{a+2}\]

how can you simplify this?

Answer 24

\[\dots\]

Answer 25

\[\frac{ \frac{ a+3 }{ (a+9)(a+2) } }{ a+2 }\]
Like this?

Answer 26

@ganeshie8 , this one is incomplete

Answer 27

the website said 1/a+9