Question

efetue as operações indicadas, simplificando a expressão

Answer 1

Resposta:

$c)\\ \\ \\(0,4x^{2}y+01x^{2}y)^{3}:((5x^{2}y).(2x^{2}y^{3})^{2}\\ \\dividendo:~desenvolva~produto~notavel:cubo~da~soma~de~dois~termos\\ \\(0,4x^{2}y)^{3}+3.(0,4x^{2}y)^{2}.0,1x^{2}y+3.0,4x^{2}y.(0,1x^{2}y)^{2}+(0,1x^{2}y)^{3}:\\  \\(5x^{2}y).(2x^{2}y^{3})^{2}\\  \\0,064x^{6}y^{3}+3.0,16x^{4}y^{2}.0,1x^{2}y+3.0,4x^{2}y.(0,1x^{2}y)^{2}+(0,1x^{2}y)^{3}:\\ \\(5x^{2}y).(2x^{2}y^{3})\\ \\0,064x^{6}y^{3}+0,048x^{6}y^{3}+0,012x^{6}y^{3}+0,01x^{6}y^{3}:(5x^{2}y).(2x^{2}y^{3})^{2}$

$\frac{0,242x^{6}y^{3}}{(5x^{2}y).(2x^{2}y^{3})^{2}}\\ \\\frac{0,242x^{6}y^{3}}{(5x^{2}y).(4x^{4}y^{6})}=\frac{0,242x^{6}y^{3}}{20x^{6}y^{7}}=\frac{0,0121}{y^{4}}$