Квадратни корен

Текст задатака објашњених у видео лекцији:

Пр.1)   Израчунати дужину странице квадрата чија је површина $81c{m^2}$.

Пр.2)   Решити једначину ${x^2} = 49$.

Пр.3)   Израчунати:

           а) $\sqrt {64}  = $

           б) $\sqrt {225}  = $

           в) $\sqrt {0,01}  = $

           г) $\sqrt {2\frac{2}{{49}}}  = $

Пр.4)   Израчунати:

           а) $\sqrt {1 + 1\frac{1}{4}}  - \sqrt {1\frac{1}{3} - 1\frac{2}{9}}  = $

           б) $\left( {\sqrt {225}  - 3\sqrt {16} } \right) \cdot \left( {2\sqrt {64}  - 9\sqrt {121} } \right) = $

           в) $\frac{3}{4}\sqrt {1\frac{7}{9}}  + \frac{1}{2}\sqrt {16}  = $

Пр.1)   $P = 81c{m^2}$

$\underline {a = ?} $

$P = {a^2}$

$81 = {a^2}$

\[\begin{array}{*{20}{c}}
\begin{gathered}
81 = {9^2} \hfill \\
\boxed{a = 9cm} \hfill \\
\end{gathered} &{}&\begin{gathered}
81 = {\left( { - 9} \right)^2} \hfill \\
a = - 9 \hfill \\
\end{gathered}
\end{array}\]

Пр.2)   

\[\begin{gathered}
{x^2} = 49 \hfill \\
\begin{array}{*{20}{c}}
{{7^2} = 49}&{}&{{{\left( { - 7} \right)}^2} = 49}
\end{array} \hfill \\
\begin{array}{*{20}{c}}
{x = 7}&{}&{x = - 7}
\end{array} \hfill \\
\hfill \\
{x^2} = 49 \hfill \\
x = \pm \sqrt {49} \hfill \\
x = \pm 7 \hfill \\
\end{gathered} \]

Пр.3)   

\[\begin{gathered}
\sqrt {64} = 8 \hfill \\
\sqrt {225} = 15 \hfill \\
\sqrt {0,01} = 0,1 \hfill \\
\sqrt {\frac{{25}}{{121}}} = \frac{5}{{11}} \hfill \\
\sqrt {2\frac{2}{{49}}} = \sqrt {\frac{{100}}{{49}}} = \frac{{\sqrt {100} }}{{\sqrt {49} }} = \frac{{10}}{7} = 1\frac{3}{7} \hfill \\
\end{gathered} \]

Пр.4)   

\[\begin{gathered}
\sqrt {169} + \frac{1}{4}\sqrt {144} - 3\sqrt {\frac{4}{9}} = 13 + \frac{1}{4} \cdot 12 - 3 \cdot \frac{2}{3} = 13 + 3 - 2 = 14 \hfill \\
\hfill \\
\sqrt {1 + 1\frac{1}{4}} - \sqrt {1\frac{1}{3} - 1\frac{2}{9}} = \sqrt {2\frac{1}{4}} - \sqrt {1\frac{3}{9} - 1\frac{2}{9}} = \sqrt {\frac{9}{4}} - \sqrt {\frac{1}{9}} = \frac{3}{2} - \frac{1}{3} = \hfill \\
= \frac{9}{6} - \frac{2}{6} = \frac{7}{6} = 1\frac{1}{6} \hfill \\
\hfill \\
\left( {\sqrt {225} - 3\sqrt {16} } \right) \cdot \left( {2\sqrt {64} - 9\sqrt {121} } \right) = \left( {15 - 3 \cdot 4} \right) \cdot \left( {2 \cdot 8 - 9 \cdot 11} \right) = \hfill \\
= 3\left( { - 83} \right) = - 249 \hfill \\
\hfill \\
\frac{3}{4}\sqrt {1\frac{7}{9}} + \frac{1}{2}\sqrt {16} = \frac{3}{4} \cdot \sqrt {\frac{{16}}{9}} + \frac{1}{2} \cdot 4 = \frac{3}{4} \cdot \frac{4}{3} + 2 = 1 + 2 = 3 \hfill \\
\hfill \\
\frac{2}{5}\sqrt {6\frac{1}{4}} + \frac{1}{4}\sqrt {16} - 2\sqrt {0,04} = \frac{2}{5}\sqrt {\frac{{25}}{4}} + \frac{1}{4} \cdot 4 - 2 \cdot 0,2 = \hfill \\
= \frac{2}{5} \cdot \frac{5}{2} + 1 - 0,4 = 1,6 \hfill \\
\end{gathered} \]