[tex]\displaystyle \tt 1). \ \ \frac{\bigg(\dfrac{5}{6}a\bigg)^{-2}\cdot2a^{3}\cdot\bigg(\dfrac{4}{5}\bigg)^{0}}{(1,2a)^{2}\cdot a^{-3}}=\frac{2a^{3}\cdot a^{3}}{\bigg(\dfrac{5a}{6}\bigg)^{2}\cdot\bigg(\dfrac{6a}{5}\bigg)^{2}}=\frac{2a^{6}}{a^{4}}=2a^{2}\\\\\\{} \ \ \ \ \ \ npu \ \ a=12000000=1,2\cdot10^{7}\\\\{} \ \ \ \ \ \ 2a^{2}=2\cdot(1,2\cdot10^{7})^{2}=2,88\cdot10^{14};\\\\\\2). \ \ \frac{x}{x+3} \ \ \ \ \ \ \ \ \ x+3\neq0 \ \ \Rightarrow \ \ x\in(-\infty;-3)\cup(-3;\infty)\\\\\\{} \ \ \ \ \ \ b). \ \ \frac{1-y}{4y-2-2y^{2}}=\frac{y-1}{2y^{2}-4y+2}=\frac{y-1}{2(y-1)^{2}}=\frac{1}{2(y-1)}\\\\\\{} \ \ \ \ \ \ \ \ \ \ \ \ y-1\neq0 \ \ \Rightarrow \ \ y\in(-\infty;1)\cup(1;\infty);[/tex]
[tex]\displaystyle \tt 3). \ \ \bigg(p+1-\frac{1}{1-p}\bigg):\bigg(p-\frac{p^{2}}{p-1}\bigg)=\frac{p(1-p)+1-p-1}{1-p}:\frac{p(p-1)-p^{2}}{p-1}=\\\\\\{} \ \ \ \ \ \ =-\frac{p^{2}}{1-p}\cdot\bigg(-\frac{p-1}{p}\bigg)=-\frac{p^{2}}{1-p}\cdot\frac{1-p}{p}=-p;[/tex]
[tex]\displaystyle \tt 4). \ \ \frac{b}{a-b}-\frac{a^{3}-ab^{2}}{a^{2}+b^{2}}\cdot\bigg(\frac{a}{(a-b)^{2}}-\frac{b}{a^{2}-b^{2}}\bigg)=\\\\\\{} \ \ =\frac{b}{a-b}-\frac{a^{3}-ab^{2}}{a^{2}+b^{2}}\cdot\frac{a(a+b)-b(a-b)}{(a-b)^{2}(a+b)}=\\\\\\{} \ \ =\frac{b}{a-b}-\frac{a(a-b)(a+b)}{a^{2}+b^{2}}\cdot\frac{a^{2}+b^{2}}{(a-b)^{2}(a+b)}=\\\\\\{} \ \ =\frac{b}{a-b}-\frac{a}{a-b}=\frac{b-a}{a-b}=-\frac{a-b}{a-b}=-1;[/tex]
