Question

Срочно надо
1 и 2 ....​

Answer 1

$\displaystyle\bf\\1) \ \frac{a^{2} -9}{a^{2} -6a+9} =\frac{(a-3)(a+3)}{(a-3)^{2} } =\frac{a+3}{a-3} \\\\\\2) \ \frac{b-3}{3b} -\frac{5-b}{b^{2} } =\frac{(b-3)\cdot b-(5-b)\cdot 3}{3b^{2} }=\frac{b^{2} -3b-15+3b}{3b^{2} } =\\\\\\=\frac{b^{2}-15 }{3b^{2} }$

Answer 2

1.

$\frac{a^2-9}{a^2-6a+9}=\frac{(a-3)(a+3)}{(a^2-3)^2}=\frac{a+3}{a-3}$

Вiдповiдь:    $\frac{a+3}{a-3}$

2.

$\frac{b-3}{3b} -\frac{5-b}{b^2}= \frac{(b-3)b-(5-b)3}{3b^2} =\frac{b^2-3b-15+3b}{3b^2} =\frac{b^2-15}{3b^2}$

Вiдповiдь:    $\frac{b^2-15}{3b^2}$