Скуп реалних бројева – понављање

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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)   

$$\begin{aligned}&\frac{1}{3}{x^2} + \frac{4}{5} = \frac{{64}}{{75}} \\ &\frac{1}{3}{x^2} = \frac{{64}}{{75}} - \frac{4}{5} \\ &\frac{1}{3}{x^2} = \frac{4}{{75}} \\ &{x^2} = \frac{4}{{25}} \\ &\begin{array}{cccccccccccccccccccc} {x = \frac{2}{5}}&{}&{x = - \frac{2}{5}} \end{array}\end{aligned}$$

Пр.5)             а)

$$\begin{aligned}&\underline {P = 100c{m^2}} \\ &O = ? \\ &\begin{array}{cccccccccccccccccccc} \begin{aligned}&P = {a^2} \\ &{a^2} = 100 \\ &a = 10cm\end{aligned} &{}&\begin{aligned}&O = 4a \\ &O = 4 \cdot 10 \\ &O = 40cm\end{aligned} \end{array}\end{aligned}$$

   б)

$$\begin{aligned}&\underline {P = 256c{m^2}} \\ &O = ? \\ &\begin{array}{cccccccccccccccccccc} \begin{aligned}&P = {a^2} \\ &{a^2} = 256 \\ &a = 16cm\end{aligned} &{}&\begin{aligned}&O = 4a \\ &O = 4 \cdot 16 \\ &O = 64cm\end{aligned} \end{array}\end{aligned}$$

   в) 

$$\begin{aligned}&\underline {P = 10c{m^2}} \\ &O = ? \\ &\begin{array}{cccccccccccccccccccc} \begin{aligned}&P = {a^2} \\ &{a^2} = 10 \\ &a = \sqrt {10} cm\end{aligned} &{}&\begin{aligned}&O = 4a \\ &O = 4 \cdot \sqrt {10} \\ &O = 4\sqrt {10} cm\end{aligned} \end{array}\end{aligned}$$

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) $\sqrt {1 + {{\left( {1\frac{1}{3}} \right)}^2}} + 5\frac{1}{3}\sqrt {2 + {{\left( { - \frac{1}{2}} \right)}^2}} - \sqrt {{{\left( { - 1\frac{1}{2}} \right)}^2}} \sqrt {1\frac{9}{{25}} - 1} =$

$= \sqrt {1 + {{\left( {\frac{4}{3}} \right)}^2}} + \frac{{16}}{3}\sqrt {2 + \frac{1}{4}} - \left| { - 1\frac{1}{2}} \right|\sqrt {\frac{9}{{25}}} = \sqrt {1 + \frac{{16}}{9}} + \frac{{16}}{3}\sqrt {2\frac{1}{4}} - \frac{3}{2} \cdot \frac{3}{5} =$

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

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