Question

Me ajudem, urgente!

Primeira conta> (0,5+x) * (0,5-x) =
Segunda conta> (T³+3) * (T³-3) =
Terceira conta> (2x³+2a) * 2x³-2a =
Quarta conta> (-3a+4m²) * (-3a-4m²) =
Quinta conta> (a²c+d²) * (a²c-d²) =
Sexta conta> (mn-7) * (mn+7)

E também têm essas aqui, rsrs:
(x/4 + 2/3) * (x/4-2/3) =
(3+2x/7) * (3-2x/7) =
(3x/4-a/5) * (3x/4+a/5) =
(a+b)³ =
( a-b)³ =

Answer 1

(0,5+x) * (0,5-x) =
0,5 * 0,5 - 0,5x + 0,5x + x *(-x) =
0,25 - x²


(t³+3) * (t³-3) =
t³ *t³ - 3t³ + 3t³ + 3* (-3) =
t⁶ - 9


(2x³+2a) * (2x³-2a) =
2x³*2x³ + 2x³*(-2a) + 2a*(2x³) + 2a *(-2a) =
4x⁶ - 4x³a + 4x³a - 4a² =
4x⁶ - 4a² =
4(x⁶ - a²)


(-3a + 4m²) * (-3a - 4m²) =
(-3a)*(-3a) + (-3a)*(-4m²) + 4m² *(-3a) + 4m² *(-4m²) =
9a² + 12m²a - 12m²a - 16m⁴ =
9a² - 16m⁴


(a²c + d²) * (a²c - d²) =
(a²c)*(a²c) + a²c*(-d²) + d²*(a²c) + d² *(-d²) =
a⁴c² - a²d²c + a²d²c - d⁴ =
a⁴c² - d⁴


(mn - 7) * (mn + 7) =
(mn)*(mn) + 7mn - 7mn - 7*7 =
m²n² - 49


$( \frac{x}{4} + \frac{2}{3} ) *( \frac{x}{4} - \frac{2}{3} ) = \\ \\ (\frac{x}{4} * \frac{x}{4} ) + \frac{x}{4} *(- \frac{2}{3}) + (\frac{2}{3} * \frac{x}{4}) + \frac{2}{3} * (- \frac{2}{3} ) = \\ \\ \frac{x^{2}}{16} - \frac{2x}{12} + \frac{2x}{12} - \frac{4}{9} = \\ \\ \frac{x^{2}}{16} - \frac{4}{9}<strong> </strong>$



$(3 + \frac{2}{7} )*(3 - \frac{2}{7} ) = \\ \\ 3*3 + 3*(- \frac{2}{7} )+3*( \frac{2}{7} ) + \frac{2}{7} *(- \frac{2}{7} ) = \\ \\ 9 - \frac{6}{7} + \frac{6}{7} - \frac{4}{49} = \\ \\ ~9 - \frac{4}{49}$



$( \frac{3x}{4} - \frac{a}{5} )* ( \frac{3x}{4} + \frac{a}{5} )= \\ \\ ( \frac{3x}{4}* \frac{3x}{4} ) + ( \frac{3x}{4} * \frac{a}{5} ) +(- \frac{a}{5} * \frac{3x}{4}) + (- \frac{a}{5} * \frac{a}{5} ) = \\ \\ \frac{9x^{2}}{16} + \frac{3ax}{20} - \frac{3ax}{20} - \frac{a^{2}}{25} = \\ \\ \frac{9x^{2}}{16} - \frac{a^{2}}{25}<strong> </strong>$



(a + b)³ =
(a + b) (a + b) (a + b) =
(a² + 2ab + b²) (a + b) =
a³ + a²b + 2a²b + 2ab² + ab² + b³ =
a³ + 3a²b + 3ab² + b³


(a - b)³ =
(a - b) (a - b) (a - b) =
(a² - 2ab + b²) (a - b) =
a³ - a²b - 2a²b + 2ab² + ab² - b³ =
a³ - 3a²b + 3ab² - b³