doturitip9

Answered

2022-07-10

How to simplify an expression?

I have tried to simplify this expression for quite a long time now but I can't find how to do it.

$(\frac{1}{2+4m}-\frac{1-m}{8{m}^{3}+1}:\frac{1-2m}{2{m}^{2}-2m+1})\cdot \frac{4m+2}{2m-1}-\frac{1}{1-4m+4{m}^{2}}$

Can someone help me with it?

I have tried to simplify this expression for quite a long time now but I can't find how to do it.

$(\frac{1}{2+4m}-\frac{1-m}{8{m}^{3}+1}:\frac{1-2m}{2{m}^{2}-2m+1})\cdot \frac{4m+2}{2m-1}-\frac{1}{1-4m+4{m}^{2}}$

Can someone help me with it?

Answer & Explanation

Shawn Castaneda

Expert

2022-07-11Added 17 answers

Starting with the original expression:

$(\frac{1}{2+4m}-\frac{1-m}{8{m}^{3}+1}:\frac{1-2m}{2{m}^{2}-2m+1})\cdot \frac{4m+2}{2m-1}-\frac{1}{1-4m+4{m}^{2}}$

Factoring where possible and changing the order of some terms:

$(\frac{1}{2(2m+1)}+\frac{m-1}{(2m+1)(4{m}^{2}-2m+1)}:\frac{-(2m-1)}{2{m}^{2}-2m+1})\cdot \frac{2(2m+1)}{2m-1}-\frac{1}{(2m-1{)}^{2}}$

Converting division into multiplication by the reciprocal; distributing the outside fraction:

$(\frac{1}{2(2m+1)}\frac{2(2m+1)}{2m-1}+\frac{m-1}{(2m+1)(4{m}^{2}-2m+1)}\frac{2{m}^{2}-2m+1}{-(2m-1)}\frac{2(2m+1)}{2m-1})-\frac{1}{(2m-1{)}^{2}}$

Some cancellations:

$\phantom{\rule{-0.8mm}{0ex}}(\frac{1}{\overline{)2(2m+1)}}\frac{\overline{)2(2m+1)}}{2m-1}+\frac{m-1}{\overline{)(2m+1)}(4{m}^{2}-2m+1)}\frac{2{m}^{2}-2m+1}{-(2m-1)}\frac{2\overline{)(2m+1)}}{2m-1})-\frac{1}{(2m-1{)}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2m-1}-\frac{2(m-1)}{4{m}^{2}-2m+1}\frac{2{m}^{2}-2m+1}{(2m-1{)}^{2}}-\frac{1}{(2m-1{)}^{2}}$

Finding common denominators:

$\frac{8{m}^{3}-8{m}^{2}+4m-1}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}-\frac{4{m}^{3}-8{m}^{2}+6m-2}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}-\frac{4{m}^{2}-2m+1}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}$

Adding/subtracting:

$\frac{4{m}^{3}-4{m}^{2}}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}$

Factoring one last time:

$\frac{4{m}^{2}(m-1)}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}$

$(\frac{1}{2+4m}-\frac{1-m}{8{m}^{3}+1}:\frac{1-2m}{2{m}^{2}-2m+1})\cdot \frac{4m+2}{2m-1}-\frac{1}{1-4m+4{m}^{2}}$

Factoring where possible and changing the order of some terms:

$(\frac{1}{2(2m+1)}+\frac{m-1}{(2m+1)(4{m}^{2}-2m+1)}:\frac{-(2m-1)}{2{m}^{2}-2m+1})\cdot \frac{2(2m+1)}{2m-1}-\frac{1}{(2m-1{)}^{2}}$

Converting division into multiplication by the reciprocal; distributing the outside fraction:

$(\frac{1}{2(2m+1)}\frac{2(2m+1)}{2m-1}+\frac{m-1}{(2m+1)(4{m}^{2}-2m+1)}\frac{2{m}^{2}-2m+1}{-(2m-1)}\frac{2(2m+1)}{2m-1})-\frac{1}{(2m-1{)}^{2}}$

Some cancellations:

$\phantom{\rule{-0.8mm}{0ex}}(\frac{1}{\overline{)2(2m+1)}}\frac{\overline{)2(2m+1)}}{2m-1}+\frac{m-1}{\overline{)(2m+1)}(4{m}^{2}-2m+1)}\frac{2{m}^{2}-2m+1}{-(2m-1)}\frac{2\overline{)(2m+1)}}{2m-1})-\frac{1}{(2m-1{)}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2m-1}-\frac{2(m-1)}{4{m}^{2}-2m+1}\frac{2{m}^{2}-2m+1}{(2m-1{)}^{2}}-\frac{1}{(2m-1{)}^{2}}$

Finding common denominators:

$\frac{8{m}^{3}-8{m}^{2}+4m-1}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}-\frac{4{m}^{3}-8{m}^{2}+6m-2}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}-\frac{4{m}^{2}-2m+1}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}$

Adding/subtracting:

$\frac{4{m}^{3}-4{m}^{2}}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}$

Factoring one last time:

$\frac{4{m}^{2}(m-1)}{(4{m}^{2}-2m+1)(2m-1{)}^{2}}$

Most Popular Questions