Question

\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)

\(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}\)

\(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}\)

Answer 1

(x-6)(x+6)/2x+10 * -3(x-6)= 3x+18/2x+10

(x-3y)(x+3y)/x^2y^2* 3xy/2(x-3y)=3x+9y/2xy

3(x-y)(x+y)/5xy * -15x^2y/2(X-y)=-9x/2

Answer 2

\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x+6\right)\left(x-6\right)}{2x+10}.\frac{3}{-x+6}.\)

\(=\frac{x-6}{2x+10}.\frac{3}{-1}=\frac{3x+18}{-2x-10}\)