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Рационални алгебарски изрази 3

Дистрибутивни закон. Разлика квадрата. Растављање алгебарских израза на чиниоце. Једноставни примери.

Задаци

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Пр.5  Растави изразе на чиниоце:

           ${x^2} - 49 = $

           $9{x^2} - 64{y^2} = $

           $0,04{x^2} - {y^2} = $

           $\frac{{16{a^2}}}{{25}} - \frac{{36}}{{8 + {b^2}}} = $

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

           $36{a^2} - {\left( {b - 2a} \right)^2} = $

           ${\left( {x - y} \right)^2}-25{\left( {x + y} \right)^2} = $

Пр.5

${x^2} - 49 = {x^2} - {7^2} = \left( {x - 7} \right)\left( {x + 7} \right) $

 

$9{x^2} - 64{y^2} = {\left( {3x} \right)^2} - \left( {8{y^2}} \right) = \left( {3x - 8y} \right)\left( {3x + 8y} \right)$

 

$0,04{x^2} - {y^2} = {\left( {0,2x} \right)^2} - {\left( {{y^2}} \right)^2} = \left( {0,2x - {y^2}} \right)\left( {0,2x + {y^2}} \right)$

 

$\frac{{16{a^2}}}{{25}} - \frac{{36}}{{8 + {b^2}}} = \frac{{16{a^2}}}{{25}} - \frac{{36}}{{81{b^2}}} = {\left( {\frac{{4a}}{5}} \right)^2} - {\left( {\frac{6}{{9b}}} \right)^2} = \left( {\frac{{4a}}{5} - \frac{6}{{9b}}} \right)\left( {\frac{{4a}}{5} + \frac{6}{{9b}}} \right)$

 

${\left( {x - 3} \right)^2} - 4 = {\left( {x - 3} \right)^2} - 4 = {\left( {x - 3} \right)^2} - {2^2} = \left( {\left( {x - 3} \right) - 2} \right)\left( {\left( {x - 3} \right) + 2} \right) = $

$= \left( {x - 3 - 2} \right)\left( {x - 3 + 2} \right) = \left( {x - 5} \right)\left( {x - 1} \right)$

 

$36{a^2} - {\left( {b - 2a} \right)^2} = 36{a^2} - {\left( {b - 2a} \right)^2} = {\left( {6a} \right)^2} - {\left( {b - 2a} \right)^2} =$
$= \left( {6a - \left( {b - 2a} \right)} \right)\left( {6a + \left( {b - 2a} \right)} \right) = \left( {6a - b + 2a} \right)\left( {6a + b - 2a} \right) =$
$= \left( {8a - b} \right)\left( {4a + b} \right)$

 

${\left( {x - y} \right)^2}-25{\left( {x + y} \right)^2} = {\left( {x - y} \right)^2} - 25{\left( {x + y} \right)^2} = {\left( {x - y} \right)^2} - {\left( {5\left( {x + y} \right)} \right)^2} =$
$= \left( {\left( {x - y} \right) + \left( {5\left( {x + y} \right)} \right)} \right)\left( {\left( {x - y} \right) - \left( {5\left( {x + y} \right)} \right)} \right) =$
$= \left( {x - y + 5x + 5y} \right)\left( {x - y - 5x - 5y} \right) =$
$= \left( {6x + 4y} \right)\left( { - 4x - 6y} \right) $