Седми разред основне школе

Квадрат бинома

Дефиниција, решени задаци.

Задаци

Текст задатака објашњених у видео лекцији:

Пр.1)   Помножино следеће полиноме:

           а) $x + 6$ и $x + 6$

           б) $3x - 5$ и $3x - 5$

Пр.2)   Применом формуле за квадрат бинома израчунати:

           а) $\left( {x + 7} \right) \cdot \left( {x + 7} \right)$

           б) $\left( {5x - 1} \right) \cdot \left( {5x - 1} \right)$

           в) ${\left( {x - 2y} \right)^2}$

           г) ${\left( {1\frac{1}{3}a + b} \right)^2}$

           д) ${\left( {5{x^2} - 4} \right)^2}$

           ђ) ${\left( {{x^2} - 0,5{y^2}} \right)^2}$

Пр.3)   Решити једначине:

           а) ${\left( {x - 5} \right)^2} - x \cdot \left( {x - 4} \right) =  - 3$

           б) ${\left( {x - 6} \right)^2} - {\left( {x + 8} \right)^2} =  - 30$

Пр.4)   Ако је $A = 2x + 3,B = 3x + 2$ и $C = 2x - 3$, израчунати:

           а) ${B^2} - A \cdot C$        б) ${A^2} + {B^2} - {C^2}$

 

 

Пр.1)   

а) ${\left( {x + 6} \right)\left( {x + 6} \right) = {{\left( {x + 6} \right)}^2} = {x^2} + 2 \cdot x \cdot 6 + {6^2} = {x^2}{\text{ }} + 12 \cdot x + 36}$

б) ${\left( {3x - 5} \right)\left( {3x - 5} \right) = {{\left( {3x - 5} \right)}^2} = {{\left( {3x} \right)}^2} - 2 \cdot 3x \cdot 5 + {\text{ }}{5^2} = 9{x^2} - 30x + 25}$

 

Пр.2)   

а) $\left( {x + 7} \right) \cdot \left( {x + 7} \right) = {\left( {x + 7} \right)^2} = {x^2} + 14x + 49$

б) $\left( {5x - 1} \right) \cdot \left( {5x - 1} \right) = {\left( {5x - 1} \right)^2} = 25{x^2} - 10x + 1$

в) ${\left( {x - 2y} \right)^2} = {x^2} - 4xy + 4{y^2}$

г) ${\left( {1\frac{1}{3}a + b} \right)^2} = {\left( {\frac{4}{3}a + b} \right)^2} = \frac{{16}}{9}{a^2} + \frac{8}{3}ab + {b^2}$

д) ${\left( {5{x^2} - 4} \right)^2} = 25{x^4} - 40{x^2} + 16$

ђ) ${\left( {{x^2} - 0,5{y^2}} \right)^2} = {x^4} - {x^2}{y^2} + 0,25{y^4}$

 

Пр.3)   

а) 

\[\begin{gathered}
{\left( {x - 5} \right)^2} - x \cdot \left( {x - 4} \right) = - 3 \hfill \\
{x^2} - 10x + 25 - {x^2} + 4x = - 3 \hfill \\
- 6x = - 28 \hfill \\
x = \frac{{28}}{6} = \frac{{14}}{3} = 4\frac{2}{3} \hfill \\
\end{gathered} \]

б) 

\[\begin{gathered}
{\left( {x - 6} \right)^2} - {\left( {x + 8} \right)^2} = - 30 \hfill \\
{x^2} - 12x + 36 - \left( {{x^2} + 16x + 64} \right) = - 30 \hfill \\
{x^2} - 12x + 36 - {x^2} - 16x - 64 = - 30 \hfill \\
- 28x = - 2 \hfill \\
x = \frac{2}{{28}} = \frac{1}{{14}} \hfill \\
\end{gathered} \]

 

Пр.4)   $A = 2x + 3,B = 3x + 2$ и $C = 2x - 3$

           а)

\[\begin{gathered}
{B^2} - A \cdot C = {\left( {3x + 2} \right)^2} - \left( {2x + 3} \right)\left( {2x - 3} \right) = 9{x^2} + 12x + 4 - \left( {4{x^2} - 9} \right) = \hfill \\
= 9{x^2} + 12x + 4 - 4{x^2} + 9 = 5{x^2} + 12x + 13 \hfill \\
\end{gathered} \]

        б) 

\[\begin{gathered}
{A^2} + {B^2} - {C^2} = {\left( {2x + 3} \right)^2} + {\left( {3x + 2} \right)^2} - {\left( {2x - 3} \right)^2} = \hfill \\
= 4{x^2} + 12x + 9 + 9{x^2} + 12x + 4 - \left( {4{x^2} - 12x + 9} \right) = \hfill \\
= 4{x^2} + 12x + 9 + 9{x^2} + 12x + 4 - 4{x^2} + 12x - 9 = \hfill \\
= 9{x^2} + 36x + 4 \hfill \\
\end{gathered} \]

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