Question 200938
I'll do the first three to get you going in the right direction.



# 1


{{{4/sqrt(5)}}} Start with the given expression.



{{{(4*sqrt(5))/(sqrt(5)*sqrt(5))}}} Multiply both the numerator and denominator by {{{sqrt(5)}}}



{{{(4*sqrt(5))/(sqrt(5*5))}}} Combine the roots.



{{{(4*sqrt(5))/(sqrt(25))}}} Multiply



{{{(4*sqrt(5))/5}}} Take the square root of 25 to get 5




So {{{4/sqrt(5)=(4*sqrt(5))/5}}}



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# 2



{{{(4*sqrt(x^12))/(24*sqrt(x^12))}}} Start with the given expression.



{{{(4*sqrt(x^12))/(4*6*sqrt(x^12))}}} Factor 24 to get 4*6



{{{(cross(4)*sqrt(x^12))/(cross(4)*6*sqrt(x^12))}}} Cancel out the common terms.



{{{(sqrt(x^12))/(6*sqrt(x^12))}}} Simplify



{{{sqrt((x^12)/(x^12))/(6)}}} Combine the roots



{{{sqrt(1)/6}}} Divide



{{{1/6}}} Simplify



So {{{(4*sqrt(x^12))/(24*sqrt(x^12))=1/6}}} where {{{x>0}}}



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# 3



{{{(15*sqrt(8x^16))/(5*sqrt(2x^4))}}} Start with the given expression.



{{{(5*3*sqrt(8x^16))/(5*sqrt(2x^4))}}} Factor 15 into 5*3



{{{(cross(5)*3*sqrt(8x^16))/(cross(5)*sqrt(2x^4))}}} Cancel out the common terms.



{{{(3*sqrt(8x^16))/(sqrt(2x^4))}}} Simplify



{{{3*sqrt((8x^16)/(2x^4))}}} Combine the roots.



{{{3*sqrt((4*2x^16)/(2x^4))}}} Factor 8 into 4*2



{{{3*sqrt((4*cross(2)x^16)/(cross(2)x^4))}}} Cancel out the common terms.



{{{3*sqrt((4x^16)/(x^4))}}} Simplify



{{{3*sqrt(4x^(16-4))}}} Divide the variable terms by subtracting the exponents.



{{{3*sqrt(4x^12)}}} Subtract



{{{3*sqrt(4*x^2*x^2*x^2*x^2*x^2*x^2)}}} Factor {{{x^12}}} into {{{x^2*x^2*x^2*x^2*x^2*x^2}}} note: there are 6 {{{x^2}}} terms.



{{{3*sqrt(4)*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)}}} Break up the square root using the identity {{{sqrt(A*B)=sqrt(A)*sqrt(B)}}}.



{{{3*2*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)}}} Take the square root of {{{4}}} to get {{{2}}}.



{{{3*2*x*x*x*x*x*x}}} Take the square root of {{{x^2}}} to get {{{x}}}.



{{{6x^6}}} Multiply.


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Answer:



So {{{(15*sqrt(8x^16))/(5*sqrt(2x^4))}}} simplifies to {{{6x^6}}}



In other words, {{{(15*sqrt(8x^16))/(5*sqrt(2x^4))=6x^6}}} where {{{x>0}}}




Edit: The other poster is correct in saying that "x" doesn't need to be non-negative, BUT recall that {{{sqrt(x^2)=abs(x)}}}. If we assume that {{{x>=0}}} (ie it is NOT negative), then {{{sqrt(x^2)=x}}} which simplifies things greatly. If "x" can be any number, then you must include the absolute value.