Question

Допоможіть будь ласка алгебра

Answer 1

Ответ:

$1) \ \ -2\\\\2) \ \ \dfrac{a-b+3c}{a+b-c}$

Объяснение:

$a) \ \ \left(\dfrac{2}{a}-\dfrac{4}{a+2}\right)\times \left(2+\dfrac{a^2+4}{a-2}\right)= \\\\ \\ =\left(\dfrac{2(a+2)}{a(a+2)}-\dfrac{4a}{a(a+2)}\right)\times \left(\dfrac{2(a-2)}{a-2}+\dfrac{a^2+4}{a-2}\right)= \\\\\\ =\dfrac{2a+4-4a}{a(a+2)}\times \dfrac{2a-4+a^2+4}{a-2}=\dfrac{(4-2a) \times (a^2+2a)}{a(a+2)(a-2)}=\\\\\\=\dfrac{-2(a-2) \times a(a+2)}{a(a+2)(a-2)}=\boxed{-2}$

=======================================================

$b) \ \ \dfrac{\dfrac{a-b}{c}+3}{\dfrac{a+b}{c}-1} = \dfrac{\dfrac{a-b+3c}{c}}{\dfrac{a+b-c}{c}}= \boxed{\dfrac{a-b+3c}{a+b-c}}$