va rog , dau coroană​

[tex]\displaystyle\bf\\4)~Simplifica:\\\\a)~\frac{4+8}{2+6}=\frac{12^{(4}}{8}=\frac{3}{2}\\\\b)~\frac{2^8}{2^5}=2^{8-5}=2^3=8\\\\c)~\frac{5^7\cdot3^9\cdot2^{10}}{5^8\cdot3^{10}\cdot2^9}=5^{7-8}\cdot3^{9-10}\cdot2^{10-9}=5^{-1}\cdot3^{-1}\cdot2^{1}=\frac{2}{5\cdot3}=\frac{2}{15}\\\\d)~\frac{3^{2016}-3^{2014}}{3^{2017}-3^{2015}}=\frac{3^{2014}(3^{2}-1)}{3^{2015}(3^{2}-1)}=\frac{3^{2014}}{3^{2015}}=3^{2014-2015}=3^{-1}=\frac{1}{3}\\\\e)~\frac{4a-6b}{6a-9b}=\frac{2(2a-3b)}{3(2a-3b)}=\frac{2}{3}[/tex]

[tex]\displaystyle\bf\\f)~\frac{abcabc}{123123}=\frac{1000abc+abc}{1000\cdot123+123}=\frac{abc(1000+1)}{123(1000+1)}=\frac{abc}{123}\\\\g)~\frac{1+2+3+...+2006}{2+4+6+...4012}=\frac{1+2+3+...+2006}{2(1+2+3+...+2006)}=\frac{1}{2}[/tex]

[tex]\it a)\ \dfrac{4+8}{2+6}=\dfrac{\ 12^{(4}}{8} =\dfrac{3}{2}\\ \\ \\ b)\ \dfrac{2^8}{2^5}=2^8:2^5=2^{8-5}=2^3=8\\ \\ \\ c)\ \dfrac{5^7\cdot3^9\cdot2^{10}}{5^8\cdot3^{10}\cdot2^9}=\dfrac{5^7\cdot3^9\cdot2^{9}\cdot2}{5^7\cdot5\cdot3^9\cdot3\cdot2^{9}}=\dfrac{2}{5\cdot3}=\dfrac{2}{15}[/tex]

[tex]\it d)\ \dfrac{3^{2016}-3^{2014}}{3^{2017}-3^{2015}}=\dfrac{3^{2014}(3^2-1)}{3^{2014}(3^3-3)}=\dfrac{8^{(8}}{24}=\dfrac{1}{3}\\ \\ \\ e)\ \dfrac{4a-6b}{6a-9b}=\dfrac{2(2a-3b)}{3(2a-3b)}=\dfrac{2}{3}\\ \\ \\ f)\ \dfrac{\overline{abcabc}}{123123}=\dfrac{\overline{abc000}+\overline{abc}}{123000+123}=\dfrac{\overline{abc}(1000+1)}{123(1000+1)}=\dfrac{\overline{abc}}{123}[/tex]

[tex]\it g)\ \dfrac{1+2+3+\ ...\ +2006}{2+4+6+\ ...\ +4012}=\dfrac{1+2+3+\ ...\ +2006}{2(1+2+3+\ ...\ +2006)}=\dfrac{1}{2}[/tex]