Question 892569
{{{((3/4^2-5)-(2/5-4*2))/(2*3^3/2+2^2)}}}
Apply order of operations, applying exponents first
{{{((3/16-5)-(2/5-4*2))/(2*9/2+4)}}}
Now do one round of multiplication and division. The 3/16 operation is deferred
{{{((3/16-5)-(2/5-8))/(18/2+4)}}}
Now perform the division 18/2
{{{((3/16-5)-(2/5-8))/(9+4)}}}
Next simplify (3/16-5). {{{(3/16-(5*16/16))}}}= {{{(3/16-(80/16))}}} = {{{(-77/16)}}}
Now simplify {{{(2/5-8)}}} = {{{(2/5-8*(5/5))}}} = {{{(2/5-(40/5))}}} = {{{-38/5}}}
Putting it together
{{{((-77/16)-(-38/5))/(9+4)}}}
{{{((-77/16)+(38/5))/(13)}}}
A common denominator for the top is 5*16 = 80
{{{((-77/16)*(5/5)+(38/5)*(16/16))/(13)}}}
{{{((-385/80)+(608/80))/(13)}}}
{{{((223/80))/13}}}
******************
{{{((223/80))*(1/13)}}}
{{{(223/2480)}}}
Notice that a good solver such as wolframalpha will give you
the above result.  Try entering ((3/4^2-5)-(2/5-4*2))/(2*3^3/2+2^2) at
www.wolframalpha.com