Question

ajutatima va rog la ex 9(e.f.g)

Answer 1

e)(xy+y²)/(a-3b) : (x²-y²)/(2a-6b)=
=y(x+y)/(a-3b) × 2(a-3b)/(x-y)(x+y)=
=y/1 × 2/(x-y)=
=2y/(x-y)

f)(y-8)/(x²-4):(2y-16)/(3x-6)=
=(y-8)/(x-2)(x+2) × 3(x-2)/2(y-8)=
=1/(x+2) × 3/2=
=3/2(x+2)

g)(a²-9()/(a²+6a+9) : (3-a)/(a+3)=
=(a-3)(a+3)/(a+3)²  × (a+3)/(3-a)=
=(a-3)/(a+3)  × (a+3)/(3-a)=
=-1/1  × 1/1=-1

Answer 2

$\displaystyle 9e). \frac{xy+y^2}{a-3b} : \frac{x^2-y^2}{2a-6b}= \frac{y(x+y)}{a-3b} : \frac{(x-y)(x+y)}{2(a-3b)} = \\ \\ = \frac{y(x+y)}{a-3b} \cdot \frac{2(a-3b)}{(x-y)(x+y)} = \frac{2y}{x-y}$

[tex]\displaystyle f). \frac{y-8}{x^2-4} : \frac{2y-16}{3x-6} = \frac{y-8}{(x-2)(x+2)} : \frac{2(y-8)}{3(x-2)} = \\ \\ =\frac{y-8}{(x-2)(x+2)} \cdot \frac{3(x-2)}{2(y-8)} = \frac{3}{2(x+2)} [/tex]

$\displaystyle g). \frac{a^2-9}{a^2+6a+9} : \frac{3-a}{a+3} = \frac{(a-3)(a+3)}{(a+3)^2} : \frac{3-a}{a+3} = \\ \\ =\frac{(a-3)(a+3)}{(a+3)^2} \cdot \frac{a+3}{3-a} =-1$