Fraction: simplify [(1-x)/x] * [ {(x-1)/(sqrt[x]+1)} /1]

[doughishere]

Im a dummy this morning....what step am i missing? Answer attached.


\(\displaystyle \frac{1-x}{x}*\frac{(x-1)(\frac{1}{\sqrt{x}+1})}{1}\)

 

[Deleted member 4993]

Im a dummy this morning....what step am i missing? Answer attached.


\(\displaystyle \frac{1-x}{x}*\frac{(x-1)(\frac{1}{\sqrt{x}+1})}{1}\)

\(\displaystyle \frac{1-x}{x}*\frac{(x-1)(\frac{1}{\sqrt{x}+1})}{1}\) ← this does NOT come from the given problem.

Instead it should have been:


\(\displaystyle \frac{1-x}{x}*\frac{1}{(x-1)(\frac{1}{\sqrt{x}+1})}\)

 

[doughishere]

I think im still confused....i must be missing a really simple rule or something...the way that I understand it is the following:




\(\displaystyle \frac{\frac{1-x}{x}}{(x-1)(\frac{1}{\sqrt{x}}+1)}=\frac{1-x}{x}\div\frac{1}{(x-1)(\frac{1}{\sqrt{x}}+1)}=\frac{1-x}{x}*\frac{(x-1)(\frac{1}{\sqrt{x}}+1)}{1}=.....\)


the rule being similar to:

I guess i dont see the answer right away...do i just need to not be lazy and multiply it out?

 

[gregk]

I think im still confused....i must be missing a really simple rule or something...the way that I understand it is the following:




\(\displaystyle \frac{\frac{1-x}{x}}{(x-1)(\frac{1}{\sqrt{x}}+1)}=\frac{1-x}{x}\div\frac{1}{(x-1)(\frac{1}{\sqrt{x}}+1)}=\frac{1-x}{x}*\frac{(x-1)(\frac{1}{\sqrt{x}}+1)}{1}=.....\)

No, not quite; you're flipping the second fraction twice. It should look like

\(\displaystyle
\displaystyle
\frac{(1-x)/x}{(x-1)\left(\frac1{\sqrt{x}} + 1\right)}
= \frac{1-x}x \div \frac{(x-1)\left(\frac1{\sqrt{x}}+1\right)}{1}
= \frac{1-x}x\cdot\frac1{(x-1)\left(\frac1{\sqrt{x}}+1\right)}.
\)

And yes this is exactly the same rule as with simple fractions.

 

[ksdhart2]

I think im still confused....i must be missing a really simple rule or something...the way that I understand it is the following:




\(\displaystyle \frac{\frac{1-x}{x}}{(x-1)(\frac{1}{\sqrt{x}}+1)}=\frac{1-x}{x}\div\frac{1}{(x-1)(\frac{1}{\sqrt{x}}+1)}=\frac{1-x}{x}*\frac{(x-1)(\frac{1}{\sqrt{x}}+1)}{1}=.....\)


the rule being similar to:

\(\displaystyle \dfrac{1}{2} \div \dfrac{1}{3} = \dfrac{1}{2} \times \dfrac{3}{1} = \dfrac{3}{2}\)

I guess i dont see the answer right away...do i just need to not be lazy and multiply it out?

You certainly could continue that way and do all of the multiplication and simplification and eventually you'd end up with the same answer, but there's a much easier way. Starting from here:

\(\displaystyle \dfrac{\dfrac{1-x}{x}}{(x-1)\left(\dfrac{1}{\sqrt{x}}+1\right)}\)

Use the known rule that:

\(\displaystyle \dfrac{\left(\dfrac{a}{b}\right)}{c}=\dfrac{a}{bc}\)

To get:

\(\displaystyle \dfrac{{1-x}}{x(x-1)\left(\dfrac{1}{\sqrt{x}}+1\right)}\)

Next divide both sides by (x - 1):

\(\displaystyle \dfrac{\dfrac{{1-x}}{x-1}}{x\left(\dfrac{1}{\sqrt{x}}+1\right)}\)

And continue...

 

[doughishere]

You certainly could continue that way and do all of the multiplication and simplification and eventually you'd end up with the same answer, but there's a much easier way. Starting from here:

\(\displaystyle \dfrac{\dfrac{1-x}{x}}{(x-1)\left(\dfrac{1}{\sqrt{x}}+1\right)}\)

Use the known rule that:

\(\displaystyle \dfrac{\left(\dfrac{a}{b}\right)}{c}=\dfrac{a}{bc}\)

To get:...

mucho gracias

 

[doughishere]

No, not quite; you're flipping the second fraction twice. It should look like

\(\displaystyle
\displaystyle
\frac{(1-x)/x}{(x-1)\left(\frac1{\sqrt{x}} + 1\right)}
= \frac{1-x}x \div \frac{(x-1)\left(\frac1{\sqrt{x}}+1\right)}{1}
= \frac{1-x}x\cdot\frac1{(x-1)\left(\frac1{\sqrt{x}}+1\right)}.
\)

And yes this is exactly the same rule as with simple fractions.

I didnt say thank you ...thank you!