Скуп реалних бројева - понављање
Скуп реалних бројева, понављање. Припема за проверу знања.
Задаци
Текст задатака објашњених у видео лекцији:
Пр.1) Израчунати:
а) $\sqrt {289} = $
б) $3\sqrt {169} - 2\sqrt {361} = $
в) ${7^2} - {\left( { - 9} \right)^2} = $
г) $\frac{{18}}{{{6^2}}} = $
д) $\frac{{ - 28}}{{\sqrt {196} }} + \sqrt {64} = $
Пр.2) Израчунати:
а) $ - {3^2} + {4^2} - {2^2} - {\left( { - 6} \right)^2} = $
б) ${\left( {\frac{1}{2}} \right)^2} - \frac{{{3^2}}}{8} + \left( { - \frac{3}{{{4^2}}}} \right) - {\left( { - \frac{1}{4}} \right)^2} = $
Пр.3) Израчунати:
а) $\left( {\sqrt {225} - 3\sqrt {16} } \right) \cdot \left( {2\sqrt {64} - 9\sqrt {121} } \right) = $
б) $\frac{3}{4}\sqrt {1\frac{7}{9}} + 0,5\sqrt {36} = $
Пр.4) Ако трећини квадрата неког броја додамо $\frac{4}{5}$, добијемо број $\frac{{64}}{{75}}$. О ком броју је реч?
Пр.5) Израчунати обим квадрата ако је његова површина:
а) $100c{m^2}$ б)$256c{m^2}$ в) $10c{m^2}$
Пр.6) Израчунати вредност израза:
а) $5\sqrt 2 + 3\sqrt 8 - \sqrt {50} - \sqrt {98} = $
б) $2\sqrt 5 - 5\sqrt 3 + \sqrt {75} + \sqrt {80} = $
Пр.7) Израчунати:
а) $\sqrt 8 + \frac{8}{{\sqrt 2 }} = $
б) $\frac{{15}}{{\sqrt 5 }} - 2\sqrt 5 = $
в) $\frac{{14}}{{\sqrt 7 }} + 14\sqrt 7 = $
Пр.8) $\sqrt {1 + {{\left( {1\frac{1}{3}} \right)}^2}} + 5\frac{1}{3}\sqrt {2 + {{\left( { - \frac{1}{2}} \right)}^2}} - \sqrt {{{\left( { - 1\frac{1}{2}} \right)}^2}} \sqrt {1\frac{9}{{25}} - 1} = $
Пр.1)
а) $\sqrt {289} = 17 $
б) $3\sqrt {169} - 2\sqrt {361} = 3 \cdot 13 - 2 \cdot 19 = 39 - 38 = 1$
в) ${7^2} - {\left( { - 9} \right)^2} = 49 - 81 = - 32$
г) $\frac{{18}}{{{6^2}}} = \frac{{18}}{{36}} = \frac{1}{2}$
д) $\frac{{ - 28}}{{\sqrt {196} }} + \sqrt {64} = \frac{{ - 28}}{{14}} + 8 = - 2 + 8 = 6$
Пр.2)
а) $ - {3^2} + {4^2} - {2^2} - {\left( { - 6} \right)^2} = - 9 + 16 - 4 - 36 = - 33$
б) ${\left( {\frac{1}{2}} \right)^2} - \frac{{{3^2}}}{8} + \left( { - \frac{3}{{{4^2}}}} \right) - {\left( { - \frac{1}{4}} \right)^2} = \frac{1}{4} - \frac{9}{8} + \left( { - \frac{3}{{16}}} \right) - \frac{1}{{16}} = - \frac{7}{8} - \frac{4}{{16}} = - \frac{9}{8} $
Пр.3)
а) $\left( {\sqrt {225} - 3\sqrt {16} } \right) \cdot \left( {2\sqrt {64} - 9\sqrt {121} } \right) = \left( {15 - 3 \cdot 4} \right)\left( {2 \cdot 8 - 9 \cdot 11} \right) = $
$ = \left( {15 - 12} \right)\left( {16 - 99} \right) = 3 \cdot \left( { - 83} \right) = - 249$
б) $\frac{3}{4}\sqrt {1\frac{7}{9}} + 0,5\sqrt {36} = \frac{3}{4}\sqrt {\frac{{16}}{9}} + 0,5 \cdot 6 = \frac{3}{4} \cdot \frac{4}{3} + 3 = 4$
Пр.4)
\[\begin{gathered}
\frac{1}{3}{x^2} + \frac{4}{5} = \frac{{64}}{{75}} \hfill \\
\frac{1}{3}{x^2} = \frac{{64}}{{75}} - \frac{4}{5} \hfill \\
\frac{1}{3}{x^2} = \frac{4}{{75}} \hfill \\
{x^2} = \frac{4}{{25}} \hfill \\
\begin{array}{*{20}{c}}
{x = \frac{2}{5}}&{}&{x = - \frac{2}{5}}
\end{array} \hfill \\
\end{gathered} \]
Пр.5) а)
\[\begin{gathered}
\underline {P = 100c{m^2}} \hfill \\
O = ? \hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
P = {a^2} \hfill \\
{a^2} = 100 \hfill \\
a = 10cm \hfill \\
\end{gathered} &{}&\begin{gathered}
O = 4a \hfill \\
O = 4 \cdot 10 \hfill \\
O = 40cm \hfill \\
\end{gathered}
\end{array} \hfill \\
\end{gathered} \]
б)
\[\begin{gathered}
\underline {P = 256c{m^2}} \hfill \\
O = ? \hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
P = {a^2} \hfill \\
{a^2} = 256 \hfill \\
a = 16cm \hfill \\
\end{gathered} &{}&\begin{gathered}
O = 4a \hfill \\
O = 4 \cdot 16 \hfill \\
O = 64cm \hfill \\
\end{gathered}
\end{array} \hfill \\
\end{gathered} \]
в)
\[\begin{gathered}
\underline {P = 10c{m^2}} \hfill \\
O = ? \hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
P = {a^2} \hfill \\
{a^2} = 10 \hfill \\
a = \sqrt {10} cm \hfill \\
\end{gathered} &{}&\begin{gathered}
O = 4a \hfill \\
O = 4 \cdot \sqrt {10} \hfill \\
O = 4\sqrt {10} cm \hfill \\
\end{gathered}
\end{array} \hfill \\
\end{gathered} \]
Пр.6) а) $5\sqrt 2 + 3\sqrt 8 - \sqrt {50} - \sqrt {98} = 5\sqrt 2 + 6\sqrt 2 - 5\sqrt 2 - 7\sqrt 2 = - \sqrt 2 $
б) $2\sqrt 5 - 5\sqrt 3 + \sqrt {75} + \sqrt {80} = 2\sqrt 5 - 5\sqrt 3 + 5\sqrt 3 + 4\sqrt 5 = 6\sqrt 5 $
Пр.7) а) $\sqrt 8 + \frac{8}{{\sqrt 2 }} = 2\sqrt 2 + \frac{8}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} = 2\sqrt 2 + 4\sqrt 2 = 6\sqrt 2 $
б) $\frac{{15}}{{\sqrt 5 }} - 2\sqrt 5 = \frac{{15}}{{\sqrt 5 }} \cdot \frac{{\sqrt 5 }}{{\sqrt 5 }} - 2\sqrt 5 = 3\sqrt 5 - 2\sqrt 5 = \sqrt 5 $
в) $\frac{{14}}{{\sqrt 7 }} + 14\sqrt 7 = \frac{{14}}{{\sqrt 7 }} \cdot \frac{{\sqrt 7 }}{{\sqrt 7 }} + 14\sqrt 7 = 2\sqrt 7 + 14\sqrt 7 = 16\sqrt 7 $
Пр.8) $\sqrt {1 + {{\left( {1\frac{1}{3}} \right)}^2}} + 5\frac{1}{3}\sqrt {2 + {{\left( { - \frac{1}{2}} \right)}^2}} - \sqrt {{{\left( { - 1\frac{1}{2}} \right)}^2}} \sqrt {1\frac{9}{{25}} - 1} = $
$ = \sqrt {1 + {{\left( {\frac{4}{3}} \right)}^2}} + \frac{{16}}{3}\sqrt {2 + \frac{1}{4}} - \left| { - 1\frac{1}{2}} \right|\sqrt {\frac{9}{{25}}} = \sqrt {1 + \frac{{16}}{9}} + \frac{{16}}{3}\sqrt {2\frac{1}{4}} - \frac{3}{2} \cdot \frac{3}{5} = $
$=\sqrt {\frac{9}{9} + \frac{{16}}{9}} + \frac{{16}}{3}\sqrt {\frac{9}{4}} - \frac{9}{{10}} = \sqrt {\frac{{25}}{9}} + \frac{{16}}{3} \cdot \frac{3}{2} - \frac{9}{{10}} =$
$= \frac{5}{3} + 8 - \frac{9}{{10}} = 8 + \frac{{50}}{{30}} - \frac{{27}}{{30}} = 8\frac{{23}}{{30}}$