Answer:
[tex]\large\boxed{if\ x\neq-3\ \wedge\ y\neq0\ \wedge\ x\neq y\ \wedge\ x\neq -y}[/tex]
second
[tex]\large\boxed{y\neq0\ \wedge\ x\neq-y\ (y\neq-x)}[/tex]
Step-by-step explanation:
We know that dividing by 0 is impossible.
Therefore, the denominator of the expression must be different from 0.
[tex]\text{For}\\\\\dfrac{(x-y)^2}{2xy+6y}\times\dfrac{4x+12}{x^2-y^2}\\\\\\2xy+6y\neq0\qquad(1)\\\\x^2-y^2\neq0\qquad(2)[/tex]
[tex](1)\\2xy+6y\neq0\qquad\text{distribute}\\\\2y(x+3)\neq0\iff2y\neq0\ \wedge\ x+3\neq0\\\\2y\neq0\qquad\text{divide both sides by 2}\\\boxed{y\neq0}\\\\x+3\neq0\qquad\text{subtract 3 from both sides}\\\boxed{x\neq-3}\\==========================\\(2)\\x^2-y^2\neq0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\(x-y)(x+y)\neq0\iff x-y\neq0\ \wedge\ x+y\neq0\\\boxed{x\neq y}\ \wedge\ \boxed{x\neq-y}[/tex]
[tex]\text{For}\\\\\dfrac{2(x-y)}{y(x+y)}\\\\y(x+y)\neq0\iff y\neq0 \wedge\ x+y\neq0\\\\ \boxed{y\neq0}\ \wedge\ \boxed{x\neq-y}[/tex]
