Помогите!очень надо!

451
$\frac{a^2-9}{2a^2}* \frac{a}{2a+6} = \frac{a^2-3^2}{2a}* \frac{1}{2(a+3)} = \\ \frac{(a-3)(a+3)}{2a}* \frac{1}{2(a+3)} = \frac{a-3}{2a}* \frac{1}{2} =\frac{a-3}{4a} \\ \\ a=1/27 \\ \frac{a-3}{4a} =\frac{1/27-3}{4*1/27}= \frac{-2 \frac{26}{27}}{4/27}= - \frac{80}{27}* \frac{27}{4}= -80/4=-20$

452
$( \frac{4y}{5x}- \frac{5x}{4y}):(4y+5x)= ( \frac{4y*4y}{5x*4y}- \frac{5x*5x}{4y*5x}):(4y+5x)= \\ ( \frac{(4y)^2}{20xy}- \frac{(5x)^2}{20xy}):(4y+5x)= \frac{(4y)^2-(5x)^2}{20xy}* \frac{1}{ (4y+5x)}= \\ \frac{(4y-5x)(4y+5x)}{20xy}* \frac{1}{ (4y+5x)}= \frac{4y-5x}{20xy}* \frac{1}{1}= \frac{4y-5x}{20xy} \\ \\ x=1/5 , y=1/8 \\ \frac{4y-5x}{20xy} =\frac{4*1/8-5*1/5}{20*1/5*1/8} =\frac{1/2-1}{4*1/1*1/8} =\frac{-1/2}{1/2} =-1$