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Решени задаци. Припрема за контролни задатак.

Задаци

Текст задатака објашњених у видео лекцији:

Пр.1) Упрости изразе:

а) ${x^8}:{x^3} = $

б) ${x^8} \cdot {x^3} = $

в) ${\left( {{x^3}} \right)^5} = $

г) $\left( {{a^7}:{a^4}} \right) \cdot {a^5} = $ 

д) $\frac{{{{\left( { - m} \right)}^4} \cdot {{\left( { - m} \right)}^6}:{m^8}}}{{{{\left( { - m} \right)}^8}:{m^6}}} = $

Пр.2) Израчунати:

а) $ - 5 + {2^3} - {4^2} = $

б) ${\left( { - 3} \right)^3} \cdot {\left( { - 5} \right)^2} = $

в) ${\left( { - 1} \right)^4}:{\left( { - 3} \right)^3} + 4:{3^3} = $

г) ${\left( { - 2} \right)^3} + {\left( {\sqrt 2 } \right)^2} - {\left( {2\sqrt 3 } \right)^2} - {3^3} = $ 

Пр.3) Израчунати:

а) ${\left( {\frac{3}{4}} \right)^5}:{\left( {\frac{3}{4}} \right)^3} \cdot \frac{3}{4} = $

б) $\frac{{{2^6} \cdot {2^4}:{2^3}}}{{{2^5}:{2^3}}} = $

Пр.4) Упростити следеће изразе:

а) $\frac{{{{\left( {{a^2}{b^3}} \right)}^2}}}{{a{b^4}}} \cdot \frac{{{a^3}b}}{{{{\left( {{a^2}{b^4}} \right)}^3}}} = $

б) ${\left( {\frac{{{{\left( {2{a^3}} \right)}^2}}}{{2{a^4}{b^3}}}} \right)^3} = $

Пр.5) Израчунати:

а) ${\left( { - \frac{6}{7}} \right)^2} \cdot {\left( { - 7} \right)^2} \cdot {\left( { - \frac{5}{6}} \right)^2} = $

б) $\frac{{{5^5} \cdot {{25}^4}}}{{{{125}^3}}} = $

Пр.6) Решити једначине:

а) ${5^x} \cdot {5^3} = 10$

б) ${3^{2x}} \cdot {3^4} = {9^7}$

в) ${8^x} \cdot {8^{2x}} \cdot {8^{3x}} = {8^{24}}$

Пр.7) Ако је $a$ страница једнакостраничног троугла чија је површина $9\sqrt 3 c{m^2}$, израчунати бројевну вредност израза $\frac{{{{\left( {{a^3}} \right)}^2} \cdot {{\left( { - a} \right)}^4}:{{\left( {{a^2}} \right)}^2}}}{{{a^6}:{{\left( {{a^4}:{a^2}} \right)}^2}}}.$

 

 

 

Пр.1) 

а) ${x^8}:{x^3} = {x^5}$

б) ${x^8} \cdot {x^3} = {x^{11}}$

в) ${\left( {{x^3}} \right)^5} = {x^{15}}$

г) $\left( {{a^7}:{a^4}} \right) \cdot {a^5} = {a^3} \cdot {a^5} = {a^8} $ 

д) $\frac{{{{\left( { - m} \right)}^4} \cdot {{\left( { - m} \right)}^6}:{m^8}}}{{{{\left( { - m} \right)}^8}:{m^6}}} = \frac{{{{\left( { - m} \right)}^{10}}:{m^8}}}{{{m^8}:{m^6}}} = \frac{{{m^{10}}:{m^8}}}{{{m^8}:{m^6}}} = \frac{{{m^2}}}{{{m^2}}} = 1$

Пр.2) 

а) $ - 5 + {2^3} - {4^2} =  - 5 + 8 - 16 =  - 13$

б) ${\left( { - 3} \right)^3} \cdot {\left( { - 5} \right)^2} =  - 27 \cdot 25 =  - 675$

в) ${\left( { - 1} \right)^4}:{\left( { - 3} \right)^3} + 4:{3^3} = 1:\left( { - 27} \right) + 4:27 =  - \frac{1}{{27}} + \frac{4}{{27}} = \frac{3}{{27}} = \frac{1}{9}$

г) ${\left( { - 2} \right)^3} + {\left( {\sqrt 2 } \right)^2} - {\left( {2\sqrt 3 } \right)^2} - {3^3} =  - 8 + 2 - 12 - 27 =  - 45$ 

Пр.3) 

а) ${\left( {\frac{3}{4}} \right)^5}:{\left( {\frac{3}{4}} \right)^3} \cdot \frac{3}{4} = {\left( {\frac{3}{4}} \right)^2} \cdot \frac{3}{4} = {\left( {\frac{3}{4}} \right)^3} = \frac{{27}}{{64}}$

б) $\frac{{{2^6} \cdot {2^4}:{2^3}}}{{{2^5}:{2^3}}} = \frac{{{2^7}}}{{{2^2}}} = {2^5} = 32 $

Пр.4) 

а) $\frac{{{{\left( {{a^2}{b^3}} \right)}^2}}}{{a{b^4}}} \cdot \frac{{{a^3}b}}{{{{\left( {{a^2}{b^4}} \right)}^3}}} = \frac{{{a^4}{b^6}}}{{a{b^4}}} \cdot \frac{{{a^3}b}}{{{a^6}{b^{12}}}} = \frac{{{a^7}{b^7}}}{{{a^7}{b^{16}}}} = \frac{1}{{{b^9}}}$

б) ${\left( {\frac{{{{\left( {2{a^3}} \right)}^2}}}{{2{a^4}{b^3}}}} \right)^3} = {\left( {\frac{{4{a^6}}}{{2{a^4}{b^3}}}} \right)^3} = {\left( {\frac{{2{a^2}}}{{{b^3}}}} \right)^3} = \frac{{8{a^6}}}{{{b^9}}} $

Пр.5) 

а) ${\left( { - \frac{6}{7}} \right)^2} \cdot {\left( { - 7} \right)^2} \cdot {\left( { - \frac{5}{6}} \right)^2} = {\left( {\left( { - \frac{6}{7}} \right) \cdot \left( { - 7} \right) \cdot \left( { - \frac{5}{6}} \right)} \right)^2} = {\left( { - 5} \right)^2} = 25 $

б) $\frac{{{5^5} \cdot {{25}^4}}}{{{{125}^3}}} = \frac{{{5^5} \cdot {{\left( {{5^2}} \right)}^4}}}{{{{\left( {{5^3}} \right)}^3}}} = \frac{{{5^5} \cdot {5^8}}}{{{5^9}}} = \frac{{{5^{13}}}}{{{5^9}}} = {5^4} = 625$

Пр.6) 

а) 

\[\begin{gathered}
{5^x} \cdot {5^3} = {5^{10}} \hfill \\
{5^x} = {5^{10}}:{5^3} \hfill \\
{5^x} = {5^7} \hfill \\
x = 7 \hfill \\
\end{gathered} \]

б) 

\[\begin{gathered}
{3^{2x}} \cdot {3^4} = {9^7} \hfill \\
{3^{2x + 4}} = {\left( {{3^2}} \right)^7} \hfill \\
{3^{2x + 4}} = {3^{14}} \hfill \\
2x + 4 = 14 \hfill \\
2x = 14 - 4 \hfill \\
2x = 10 \hfill \\
x = 5 \hfill \\
\end{gathered} \]

в) 

\[\begin{gathered}
{8^x} \cdot {8^{2x}} \cdot {8^{3x}} = {8^{24}} \hfill \\
{8^{x + 2x + 3x}} = {8^{24}} \hfill \\
{8^{6x}} = {8^{24}} \hfill \\
6x = 24 \hfill \\
x = 24:6 \hfill \\
x = 4 \hfill \\
\end{gathered} \]

Пр.7) 

\[\begin{gathered}
\underline {P = 9\sqrt 3 c{m^2}} \hfill \\
a = ? \hfill \\
\frac{{{{\left( {{a^3}} \right)}^2} \cdot {{\left( { - a} \right)}^4}:{{\left( {{a^2}} \right)}^2}}}{{{a^6}:{{\left( {{a^4}:{a^2}} \right)}^2}}} = ? \hfill \\
\hfill \\
P = \frac{{{a^2}\sqrt 3 }}{4} \hfill \\
9\sqrt 3 = \frac{{{a^2}\sqrt 3 }}{4} \hfill \\
{a^2}\sqrt 3 = 36\sqrt 3 \hfill \\
{a^2} = 36 \hfill \\
a = 6cm \hfill \\
\frac{{{{\left( {{a^3}} \right)}^2} \cdot {{\left( { - a} \right)}^4}:{{\left( {{a^2}} \right)}^2}}}{{{a^6}:{{\left( {{a^4}:{a^2}} \right)}^2}}} = \frac{{{a^6} \cdot {a^4}:{a^4}}}{{{a^6}:{{\left( {{a^2}} \right)}^2}}} = \frac{{{a^{10}}:{a^4}}}{{{a^6}:{a^4}}} = \frac{{{a^6}}}{{{a^2}}} = {a^4} \hfill \\
{a^4} = {6^4} = 1296 \hfill \\
\end{gathered} \]