Question

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Answer 1

Respuesta:

Explicación paso a paso:

$c.7\\d.3\\e.15\\\\2.\\d.\left(\frac{1}{2}m^{3}-\frac{2}{3}m^{2}\right)+\left(-\frac{2}{5}m^{3}+\frac{1}{6}m^{2}\right)=\\\\\texttt{Eliminando los signos de agrupaci\'on y reagrupando}\\\texttt{los t\'erminos semejantes}=\frac{1}{2}m^{3}-\frac{2}{3}m^{2}-\frac{2}{5}m^{3}+\frac{1}{2}m^{2}=\\\\=\frac{1}{2}m^{3}-\frac{2}{5}m^{3}-\frac{2}{3}m^{2}+\frac{1}{2}m^{2}=\\\\\texttt{Simplificando:}=m^{3}\left(\frac{1}{2}-\frac{2}{5}\right)+m^{2}\left(\frac{1}{2}-\frac{2}{3}\right)=\\\\=\frac{1}{10}m^{3}-\frac{1}{6}m^{2}\\\\3.-\\d.(9x-7y+9z)-(3z-4y+2x)=9x-7y+9z-3z+4y-2x=\\\\=9x-2x-7y+4y+9z-3z=7x-3y+6z\\\\e.\\(-6ab^{2}+8a^{2}b)-\left(\frac{2}{5}ab-\frac{4}{3}a^{2}b\right)=-6ab^{2}+8a^{2}b-\frac{2}{5}ab+\frac{4}{3}a^{2}b=\\\\=-6ab^{2}+8a^{2}b+\frac{4}{3}a^{2}b-\frac{2}{5}ab=-6ab^{2}+\frac{28}{3}a^{2}b-\frac{2}{5}ab=\\\\=\frac{28}{3}a^{2}b-\frac{2}{5}ab-6ab^{2}$