Дефиниције и решени задаци.
Задаци које смо решили у видео лекцији:
1. Упростити изразе:
(а) ${\left( {{x^3}} \right)^4} = $
(б) ${x^6} \cdot x = $
(в) ${x^{16}}:{x^5} = $
(г) ${\left( {{x^3}} \right)^4} \cdot {\left( {{x^2}} \right)^5} = $
2. Израчунати вредност алгебарског израза:
(а) ${x^3} + {x^2}$ за $x = 2$
(б) ${x^3} + 4{x^2} - 7x - 11$ за $x = - 2$
(в) $8{x^3} + 4{x^2} - 7x - 11$ за $x = - \frac{1}{2}$
3. Израчунати $\frac{{{2^7} \cdot {4^5}}}{{{8^8}:{{32}^2}}} = $
4. Дати су полиноми $P = {x^3} + 4{x^2} - 7x - 11$ и $Q = - 2{x^3} + 9{x^2} - 7x - 19$. Одредити њихов збир и њихову разлику.
5. Решити једначине:
(а) ${2^{3x - 1}}:{2^{4x + 5}} = {16^7}$
(б) $3x - \left( {\left( {2 + 4x} \right) + 5} \right) + 9x = - 18$
(в) ${3^{5x - 4}} \cdot {3^{2 - 3x}} = {27^5}$
1.
(а) ${\left( {{x^3}} \right)^4} = {x^{3 \cdot 4}} = {x^{12}}$
(б) ${x^6} \cdot x = {x^{6 + 1}} = {x^7}$
(в) ${x^{16}}:{x^5} = {x^{16 - 5}} = {x^{11}}$
(г) ${\left( {{x^3}} \right)^4} \cdot {\left( {{x^2}} \right)^5} = {x^{12}} \cdot {x^{10}} = {x^{22}}$
2. (а) ${x^3} + {x^2}$ за $x = 2$
\[{x^3} + {x^2} = {2^3} + {2^2} = 8 + 4 = 12\]
(б) ${x^3} + 4{x^2} - 7x - 11$ за $x = - 2$
\[{x^3} + 4{x^2} - 7x - 11 = {\left( { - 2} \right)^3} + 4{\left( { - 2} \right)^2} - 7\left( { - 2} \right) - 11 = - 8 + 16 + 14 - 11 = 11\]
(в) $8{x^3} + 4{x^2} - 7x - 11$ за $x = - \frac{1}{2}$
\[\begin{gathered}
8{x^3} + 4{x^2} - 7x - 11 = 8{\left( { - \frac{1}{2}} \right)^3} + 4{\left( { - \frac{1}{2}} \right)^2} - 7\left( { - \frac{1}{2}} \right) - 11 = \hfill \\
= 8\left( { - \frac{1}{8}} \right) + 4\left( {\frac{1}{4}} \right) - 7\left( { - \frac{1}{2}} \right) - 11 = - 1 + 1 + 3,5 - 11 = - 7,5 \hfill \\
\end{gathered} \]
3. $\frac{{{2^7} \cdot {4^5}}}{{{8^8}:{{32}^2}}} = \frac{{{2^7} \cdot {{\left( {{2^2}} \right)}^5}}}{{{{\left( {{2^3}} \right)}^8}:{{\left( {{2^5}} \right)}^2}}} = \frac{{{2^7} \cdot {2^{10}}}}{{{2^{24}}:{2^{10}}}} = \frac{{{2^{17}}}}{{{2^{14}}}} = {2^3} = 8$
4. $P = {x^3} + 4{x^2} - 7x - 11$
$Q = - 2{x^3} + 9{x^2} - 7x - 19$.
\[\begin{gathered}
P + Q = {x^3} + 4{x^2} - 7x - 11 - 2{x^3} + 9{x^2} - 7x - 19 = \hfill \\
= - {x^3} + 13{x^2} - 14x - 30 \hfill \\
P - Q = {x^3} + 4{x^2} - 7x - 11 - \left( { - 2{x^3} + 9{x^2} - 7x - 19} \right) = \hfill \\
= {x^3} + 4{x^2} - 7x - 11 + 2{x^3} - 9{x^2} + 7x + 19 = \hfill \\
= 3{x^3} - 5{x^2} + 8 \hfill \\
\end{gathered} \]
5.(а)
\[\begin{gathered}
{2^{3x - 1}}:{2^{4x + 5}} = {16^7} \hfill \\
{2^{3x - 1 - \left( {4x + 5} \right)}} = {\left( {{2^4}} \right)^7} \hfill \\
{2^{3x - 1 - 4x - 5}} = {2^{28}} \hfill \\
{2^{ - x - 6}} = {2^{28}} \hfill \\
- x - 6 = 28 \hfill \\
- x = 28 + 6 \hfill \\
- x = 34 \hfill \\
x = - 34 \hfill \\
\end{gathered} \]
(б)
\[\begin{gathered}
3x - \left( {\left( {2 + 4x} \right) + 5} \right) + 9x = - 18 \hfill \\
3x - \left( {2 + 4x + 5} \right) + 9x = - 18 \hfill \\
3x - 2 - 4x - 5 + 9x = - 18 \hfill \\
8x - 7 = - 18 \hfill \\
8x = - 11 \hfill \\
x = - \frac{{11}}{8} \hfill \\
\end{gathered} \]
(в)
\[\begin{gathered}
{3^{5x - 4}} \cdot {3^{2 - 3x}} = {27^5} \hfill \\
{3^{5x - 4 + 2 - 3x}} = {\left( {{3^3}} \right)^5} \hfill \\
{3^{2x - 2}} = {3^{15}} \hfill \\
2x - 2 = 15 \hfill \\
2x = 2 + 15 \hfill \\
2x = 17 \hfill \\
x = \frac{{17}}{2} \hfill \\
\end{gathered} \]