Question 74779
# 18) 3x^5/15x^12=1/5x^7
# 19) 3x/18-x/18=(3x-x)/18=2x/18=x/9
#20) [(x^2-49y^2)/(6x^2+42y^2)]/(x^2-7xy)
[(x-7y)(x+7y)/6(x^2+7y^2)]/x(x-7y)
[(x-7y)(x+7y)/6(x^2+7y^2)]/x(x-7y) now invert the denominator term & multiply
[(x-7y)(x+7y)x(x-7y)]/6(x^2+7y^2)
x(x-7y)(x+7y)(x-7y)/6(x^2+7y^2)