Question

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

Answer 1

Answer:

we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$

Step-by-step explanation:

Given the expression

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

$=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}$

$=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}$

$\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}$

$=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}$

cancel the common factor: 2

$=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}$

cancel the common factor: p

$=\frac{2p+1}{p^2\left(4p^2-1\right)}$

$=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}$

cancel the common factor: 2p+1

$=\frac{1}{p^2\left(2p-1\right)}$

Expanding

$=\frac{1}{2p^3-p^2}$

Thus, we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$