Пресметнете логаритмичните и изразите с корени

[publicenemy94]

Имам да реша няколко задачи с логаритмуване,коренуване... ако може малко помощ по някоя или по всичките..
http://store.picbg.net/pubpic/90/4E/a6a89f9b0d9b904e.JPG

[ammornil]

(a)
[tex]\frac{(3,(2)-0,333...)^{^{-1}}}{2^5-6}+\frac{5^{^{-1}}.5^{^{\frac{2}{3}}}}{5^{^{-\frac{2}{3}}}.5^{^{-2}}}=\frac{(3\frac{2}{9}-\frac{1}{3})^{^{-1}}}{32-6}+\frac{5^{^{(-1+\frac{2}{3})}}}{5^{^{(-\frac{2}{3}-2)}}}=\frac{(\frac{29}{9}-\frac{3}{9})^{^{-1}}}{26}+5^{^{-\frac{1}{3}-(-2\frac{2}{3})}}=\frac{(\frac{26}{9})^{^{-1}}}{26}+5^{^{\frac{7}{3}}}=9+5^{^{\frac{7}{3}}}= \\
\vspace{5} \\
=9+\sqrt[3]{5^{7}}[/tex]

(б)
[tex](\sqrt[4]{\sqrt{33}+7}.\sqrt[4]{7-\sqrt{33}}+\sqrt[5]{-32}).\sqrt{24}.\sqrt{200}=(\sqrt[4]{(7+\sqrt{33}).(7-\sqrt{33})}+\sqrt[5]){(-2)^{^{5}}}}.\sqrt{24.200}= \\
\vspace{5} \\
=(\sqrt[4]{7^{^{2}}-(\sqrt{33})^{^{2}}}-2).\sqrt{2^{^{3}}.3.2^{^{3}}.5^{^{2}}}=(\sqrt[4]{49-33}-2).40\sqrt{3}=(\sqrt[4]{2^{^{4}}}-2).40\sqrt{3}=(2-2).40\sqrt{3}=0[/tex]

[ammornil]

(в)
[tex]\frac{196^2-154^2}{7^2-28^2}=\frac{(196-154).(196+154)}{(7-28).(7+28)}=\frac{42.350}{-21.35}=-\frac{2.\cancel{21}.\cancel{35}.10}{\cancel{21}.\cancel{35}}=-20[/tex]

(г)
[tex]\frac{3^{^{\sqrt{3}-1}}}{3^{^{\sqrt{3}+1}}}=3^{^{\sqrt{3}-1-(\sqrt{3}+1)}}=3^{^{\sqrt{3}-1-\sqrt{3}-1}}=3^{^{-2}}=\frac{1}{9}[/tex]

[ammornil]

(д)
[tex]log_{_{3}} \hspace{2}729 -log_{_{2}} \hspace{2}4 -log_{_{2}} \hspace{4}256=log_{_{3}} \hspace{2}3^6-log_{_{2}} \hspace{2}2^2-log_{_{4}} \hspace{2}4^4=6-2-4=0[/tex]

(e)
[tex]log_{_{2a+3}} \hspace{2}\frac{1}{4}, \hspace{12} a=-\frac{1}{2} \\
\vspace{5} \\
log_{_{2.(-\frac{1}{2})+3}} \hspace{2} \frac{1}{4}=log_{_{2}} \hspace{2} 2^{^{-2}}=-2[/tex]

[ammornil]

(a)
[tex]\frac{(3,(2)-0,333...)^{^{-1}}}{2^5-6}+\frac{5^{^{-1}}.5^{^{\frac{2}{3}}}}{5^{^{-\frac{2}{3}}}.5^{^{-2}}}=\frac{(3\frac{2}{9}-\frac{1}{3})^{^{-1}}}{32-6}+\frac{5^{^{(-1+\frac{2}{3})}}}{5^{^{(-\frac{2}{3}-2)}}}=\frac{(\frac{29}{9}-\frac{3}{9})^{^{-1}}}{26}+5^{^{-\frac{1}{3}-(-2\frac{2}{3})}}=\frac{(\frac{26}{9})^{^{-1}}}{26}+5^{^{\frac{7}{3}}}=\frac{9}{26^2}+5^{^{\frac{7}{3}}}= \\
\vspace{5} \\
=\frac{9}{676}+\sqrt[3]{5^{7}}=\frac{9}{676}+25\sqrt[3]{5}[/tex]

[publicenemy94]

Идеални решения, мерси