Odpowiedź :
Odpowiedź:
zad 7
a)
1/5 * [10x(6y + 5)] + 1/2 * (8x + 6y) = 2x(6y + 5) + 4x + 3y =
= 12xy + 10x + 4x + 3y = 12xy + 14x + 3y
b)
x(2x - 3y + 4) - 5y(4x + 3y - 2) = 2x² - 3xy + 4x - 20xy - 15y² + 10y =
= 2x² - 15y² - 23xy + 4x + 10y
[tex]a) \: \frac{1}{5} \times (10x(6y + 5)) + \frac{1}{2} \times (8x + 6y) = \frac{1}{5} \times (60xy + 50x) + \frac{1}{2} \times 2(4x + 3y) = \frac{1}{5} \times 5(12xy + 10x) + (4x + 3y) = (12xy + 10x) + 4x + 3y = 12xy + 10x + 4x + 3y = 12xy + 14x + 3y[/tex]
[tex]b)x(2x - 3y + 4) - 5y(4x + 3y - 2) = 4x - 6y + 8 - 5y \times (4x + 3y - 2) = 4x - 6y + 8 - 20xy - 15y {}^{2} + 10y = 4x + 4y + 8 - 20xy - 15y {}^{2} [/tex]