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$1.$
$a) \frac{15xy^4}{10x^3y^2} = \frac{3y^2}{2x^2}$

$b) \frac{ab-b}{b^2}= \frac{b(a-1)}{b^2}= \frac{a-1}{b}$

$B) \frac{4x^2-y^2}{2x-y} = \frac{(2x-y)(2x+y)}{(2x-y)}= \frac{2x+y}{1} = 2x+y$
$2.$

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

$b) \frac{2x^2}{x^2-4} - \frac{2x}{x+2}= \frac{2x^2}{(x-2)(x+2)} - \frac{2x(x-2)}{(x-2)(x+2)} = \frac{2x^2-2x^2+4x}{x^2-4}=\frac{4x}{(x-2)(x+2)}$

$B) \frac{7a}{a-b}-7 = \frac{7a-7a+7b}{a-b} = \frac{7b}{a-b}$
$3.$
$\frac{5}{(a+2)^2}-\frac{5}{a^2-4}-\frac{5}{a+2}= \frac{5}{(a+2)(a+2)} - \frac{5}{(a-2)(a+2)} - \frac{5}{a+2}=$ $\frac{(5a-10)-(5a+10)-(5(a^2-4))}{(a+2)(a+2)(a-2)} = \frac{-20-5a^2+20}{(a+2)(a+2)(a-2)} = - \frac{5a^2}{(a+2)(a+2)(a-2)}$
$4.$
   $\frac{2a-2c+ax-cx}{x^2-4} = \frac{2(a-c)+x(a-c)}{(x+2)(x-2)} = \frac{(2+x)(a-c)}{(x+2)(x-2)}$
если, $a=6.7;c=5.3;x=1.9$ , то $\frac{(2+1.9)(6.7-5.3)}{(1.9+2)(1.9-2)} = \frac{3.9*1.4}{3.9*(-0.1)} = \frac{1.4}{-0,1} =- \frac{14}{-1} =-14$
$5.$
   $\frac{(x+2)^2}{2} - \frac{x^2-4}{4}- \frac{(x-2)^2}{8} = \frac{x^2}{8}$    $|*8$
   $4(x+2)(x+2)-2(x+2)(x-2)-(x-2)(x-2)=x^2$
   $20x=-20$
   $x=-1$
Ответ: -1