Question

помогите пж алгебра ​

Answer 1

$\bigg(\dfrac{(a+3b)^2}{4(a-3b)}-4a\bigg):\dfrac{3a+9b}{24a-16}\cdot\dfrac{a^2-9b^2}{5a^2-18ab-3b^2}=$

$=\dfrac{a^2+6ab+9b^2-16a(a-3b)}{4(a-3b)}\cdot\dfrac{4(6a-4)}{3(a+3b)}\cdot\dfrac{(a-3b)(a+3b)}{5a^2-18ab-3b^2}=$

$=\dfrac{a^2+6ab+9b^2-16a^2+48ab}{1}\cdot\dfrac{6a-4}{3}\cdot\dfrac{1}{5a^2-18ab-3b^2}=$

$=-15a^2+54ab+9b^2\cdot\dfrac{1}{5a^2-18ab-3b^2}\cdot\dfrac{6a-4}3=$

$=-3(5a^2-18ab-3b^2)\cdot\dfrac1{5a^2-18ab-3b^2}\cdot\dfrac{6a-4}3=-3\cdot\dfrac{6a-4}3=-(6a-4)=$

$=4-6a$

При  a = -2/3 ;  b = 7/3

$4-6\bigg(-\dfrac23\bigg)=4-(-2\cdot2)=4+4=8$

Ответ:  8