Respuesta:
Explicación paso a paso:
[tex](2y-1)^{5} = (2y)^{5} -5(2y)^{4} (1)^{1} +\frac{5*4}{2} (2y)^{3}(1)^{2}-\frac{10*3}{3} (2y)^{2} (1)^{3} +\frac{10*2}{4} (2y)^{1} (1)^{4} -(1)^{5}[/tex]
[tex]= 32y^{5} -5(16y^{4} )(1) + 10(8y^{3} )(1)-10(4y^{2}) (1)+5(2y)(1)-1[/tex]
[tex]= 32y^{5} -80y^{4} +80y^{3} -40y^{2} +10y-1[/tex]
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Respuesta:
Explicación paso a paso:
[tex](2y-1)^{5} = (2y)^{5} -5(2y)^{4} (1)^{1} +\frac{5*4}{2} (2y)^{3}(1)^{2}-\frac{10*3}{3} (2y)^{2} (1)^{3} +\frac{10*2}{4} (2y)^{1} (1)^{4} -(1)^{5}[/tex]
[tex]= 32y^{5} -5(16y^{4} )(1) + 10(8y^{3} )(1)-10(4y^{2}) (1)+5(2y)(1)-1[/tex]
[tex]= 32y^{5} -80y^{4} +80y^{3} -40y^{2} +10y-1[/tex]
Respuesta:
Explicación paso a paso: