Question

Me ajudem nessa 6 questões pfvr, e pra hoje

Answer 1

Resposta:

Explicação passo a passo:

$1) a) 20a^{2} x^{3} + 32a^{2}x^{2} : 4a^{2} \\20a^{2} x^{3} : 4a^{2} x = 5x^{2} \\32a^{3} x^{2} : 4a^{2} = 8ax\\5x^{2} + 8ax\\\\b) 15y^{6} - 33y^{5} - 21y^{2} : 3y^{2} \\15y^{6} : 3y^{2} = 5y^{4} \\- 33y^{5} : 3y^{2} = - 11 y^{3} \\- 21y^{2} : 3y^{2} = - 7\\5y^{4} - 11 y^{3} - 7\\\\\2) 25x^{6} - 4x^{4} + 30x^{3} - 35x^{2} : 5x^{2} \\25x^{6} : 5x^{2} = 5x^{4} \\- 4x^{4} : 5x^{2} = \frac{-4}{5} x^{2} \\+ 30x^{3} : 5x^{2} = 6x\\- 35x^{2} : 5x^{2} = - 7\\5x^{4} - \frac{4}{5} x^{2} + 6x +$$\\\\\\3) 7x^{3} y^{2} - x^{2} y^{2} : - 3x^{2} y = \\ 7x^{3} y^{2} : - 3x^{2} y = - \frac{7}{3} xy\\ - x^{2} y^{2} : - 3x^{2} y = +\frac{1}{3} y\\ - \frac{7}{3} xy - \frac{1}{3} y\\\\4) a) (3x + 2)^{3} (3x)^{2} + 2 . 3x . 2 + (2)^{2} \\ 9x^{2} + 12x + 4\\\\ b) (2m - n)^{2} \\ (2m)^{2} - 2 . 2m . n + (n)^{2} \\ 2m^{2} - 4mn + n^{2} \\\\5) (2x^{3} - 1) . (2x^{3} + 1)\\ (2x^{3} )^{2} - (1)^{2} \\ 4x^{6} - 1$$6) (2p + r)^{3} \\ (2p)^{3} + 3 . (2p)^{2} . r + 3 . 2p . (r)^{2} + (r)^{3} \\ 8p^{3} + 12pr + 6pr^{2} + (r)^{3} \\\\b) (4h - 1)^{3} \\ (4h)^{3} - 3 . (4h)^{2} . 1 + 3 . 4h - (1)^{3} \\ 64h3 - 48h + 12h - 1$