Answer:
【Answer】: 1/(3*(x+1))
【Explanation】: Given function is x^2/(x^2+2x+1)÷3x/(x^2-1), its numerator can be simplified as:
x^2 = x * x
As for its denominator (x^2+2x+1), completing squares:
x^2+2x+1 = (x + 1)^2.
Finally, we get above fraction as x^2/(x + 1)^2.
The second half of the division – the fraction 3x/(x^2 - 1). Here its numerator is:
3x.
The denominator (x^2 - 1) are difference of squares:
x^2 - 1 = (x + 1) * (x - 1)
Hence 3x are being divide upon two factors (violating communality law of multiplication) thus we see as 3x/((x + 1) * (x - 1)).
Given expression simplifies to
x^2/(x + 1)^2 ÷ 3x/((x + 1) * (x - 1))
Division operation is inverse of multiplication, wherein the order of numerator’s and denominator’s simplify operation while reversing the faction being divided, hence expression whence
x^2/(x + 1)^2 * ((x + 1) * (x - 1))/3x.
Further simplifying the equivalent expression (using communality law of multiplication, redrive the denominator increasing numerator in terms with factor)
[x^2 * ((x + 1) * (x - 1))]/[(x + 1)^2 * 3x]
The factor x^2 and x are reduced having 1 in numerator while functor (x + 1)^2 and (x + 1) having (x + 1) being left in denominator. Our solving results as:
1/[(x + 1) * 3]
