Ответ:

[tex]\displaystyle \boxed{4.}~~~a)16;~~b)\frac{1}{b^8}; \\ \boxed{5.}~~~\frac{2}{3} ;\\ \boxed{6.}~~~-2436;\\ \boxed{7.}~~~x=2[/tex]

Объяснение:

[tex]\displaystyle \boxed{4.}[/tex]

[tex]\displaystyle a) (a^{-3})^4*(a^{-7})^{-2}:a^{-2}=a^{-3*4}*a^{-7*(-2)}:a^{-2}=a^{-12+14+2}=a^4=16;[/tex]

[tex]\displaystyle b)(-0,4ab^{-2})^{-2}*(0,8ab^{-2})^2=\frac{1}{(-0,4ab^{-2})^2}*\frac{0,64a^2}{b^4} =\frac{1}{\frac{0,64a^2}{b^4} } *\frac{0,64a^2}{b^4} =\frac{1}{0,64a^2} *\frac{1}{b^4} *\frac{0,64a^2}{b^4} =\frac{1}{b^{4+4}} =\frac{1}{b^8}[/tex]

[tex]\displaystyle \boxed{5.}[/tex]

[tex]\displaystyle \frac{6^n}{3^{n+1}*0,5^{1-n}} =\frac{6^n}{3^n*3*0,5*\frac{1}{0,5^n} } =\frac{\not6^n}{1,5*\not6^n} =\frac{1}{1,5} =\frac{2}{3}[/tex]

[tex]\displaystyle \boxed{6.}[/tex]

[tex]\displaystyle (a^{-1}+b^{-1})(a^{-3}-a^{-2}b^{-1}+a^{-1}b^{-2}-b^{-3})=\bigg(\frac{1^{(b}}{a} +\frac{1^{(a}}{b} \bigg)\bigg(\frac{1^{(b^3}}{a^3} -\frac{1^{(ab^2}}{a^2b} +\frac{1^{(a^2b}}{ab^2} -\frac{1^{(a^3}}{b^3} \bigg)=\bigg(\frac{b+a}{ab} \bigg)\bigg(\frac{b^3-ab^2+a^2b-a^3}{a^3b^3} \bigg)=\frac{b^4-ab^3+a^2b^2-a^3b+ab^3-a^2b^2+a^3b-a^4}{a^4b^4} =\frac{b^4-a^4}{a^4b^4} =\frac{\bigg(\frac{1}{5} \bigg)^4-\bigg(\frac{1}{2}\bigg)^4 }{\bigg(\frac{1}{2}\bigg) ^4*\bigg(\frac{1}{5} \bigg)^4} =[/tex]

[tex]\displaystyle =\frac{\frac{1^{(16}}{625} -\frac{1^{(625}}{16} }{\frac{1}{64} *\frac{1}{625} } =\frac{-\frac{609}{10000} }{\frac{1}{40000} } =-\frac{609}{10000} *\frac{40000}{1} =-609*4=-2436[/tex]

[tex]\displaystyle \boxed{7.}[/tex]  задача сводится к решению показательного уравнения:

[tex]\displaystyle \frac{2^{x+2}*3^{2x}}{6^{2x}} =1;\\\frac{2^{x+2}*\not3^{2x}}{2^{2x}*\not3^{2x}} =1;\\2^{x+2}=2^{2x};\\x+2=2x;\\x=2[/tex]