Question

I need help asap someone please help me i dont understand this question so can someone help me

Answer 1

we conclude that:

`(2p)/(4p^2-1)/ (6p^3)/(6p+3)=(1)/(2p^3-p^2)`

Explanation:

Given the expression

`(2p)/(4p^2-1)/ (6p^3)/(6p+3)`

`\mathrm{Apply\:the\:fraction\:rule}:\quad (a)/(b)/ (c)/(d)=(a)/(b)* (d)/(c)`

`=(2p)/(4p^2-1)* (6p+3)/(6p^3)`

`=(2p)/(4p^2-1)* (2p+1)/(2p^3)`

`\mathrm{Multiply\:fractions}:\quad (a)/(b)* (c)/(d)=(a\:* \:c)/(b\:* \:d)`

`=(2p\left(2p+1\right))/(\left(4p^2-1\right)* \:2p^3)`

cancel the common factor: 2

`=(p\left(2p+1\right))/(\left(4p^2-1\right)p^3)`

cancel the common factor: p

`=(2p+1)/(p^2\left(4p^2-1\right))`

`=(2p+1)/(p^2\left(2p+1\right)\left(2p-1\right))`

cancel the common factor: 2p+1

`=(1)/(p^2\left(2p-1\right))`

Expanding

`=(1)/(2p^3-p^2)`

Thus, we conclude that:

`(2p)/(4p^2-1)/ (6p^3)/(6p+3)=(1)/(2p^3-p^2)`