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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) = \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}  = $

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) $ - {3^2} + {4^2} - {2^2} - {\left( { - 6} \right)^2} =  - 9 + 16 - 4 - 36 =  - 33$

           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} = $

$=\sqrt {\frac{9}{9} + \frac{{16}}{9}}  + \frac{{16}}{3}\sqrt {\frac{9}{4}}  - \frac{9}{{10}} = \sqrt {\frac{{25}}{9}}  + \frac{{16}}{3} \cdot \frac{3}{2} - \frac{9}{{10}} =$

$= \frac{5}{3} + 8 - \frac{9}{{10}} = 8 + \frac{{50}}{{30}} - \frac{{27}}{{30}} = 8\frac{{23}}{{30}}$