Question 966838
Use parentheses, they're free and avoid confusion.
I'm assuming,
{{{(9x)/(x^2+2xy-3y^2)}}} is the first term.
{{{(9x)/(x^2+2xy-3y^2)=(9x)/((x+3y)(x-y))}}}
.
.
{{{(6x)/(x^2+5xy-6y^2)=(6x)/((x+6y)(x-y))}}}
So then,
{{{(9x)/(x^2+2xy-3y^2)-(6x)/(x^2+5xy-6y^2)=(9x)/((x+3y)(x-y))-(6x)/((x+6y)(x-y))}}}
{{{(9x)/(x^2+2xy-3y^2)-(6x)/(x^2+5xy-6y^2)=((9x)(x+6y))/((x+3y)(x+6y)(x-y))-((6x)(x+3y))/((x+3y)(x+6y)(x-y))}}}
{{{(9x)/(x^2+2xy-3y^2)-(6x)/(x^2+5xy-6y^2)=((9x)(x+6y)-(6x)(x+3y))/((x+3y)(x+6y)(x-y))}}}
{{{(9x)/(x^2+2xy-3y^2)-(6x)/(x^2+5xy-6y^2)=(9x^2+54xy-6x^2-18xy)/((x+3y)(x+6y)(x-y))}}}
{{{(9x)/(x^2+2xy-3y^2)-(6x)/(x^2+5xy-6y^2)=(3x^2+36xy)/((x+3y)(x+6y)(x-y))}}}
{{{(9x)/(x^2+2xy-3y^2)-(6x)/(x^2+5xy-6y^2)=highlight_green((3x(x+12y))/((x+3y)(x+6y)(x-y)))}}}