$21a^{2}b-4b-12a+7a{b}^2=(21a^{2}b+7a{b}^2)-(12a+4b)=7ab(3a+b)-4(3a+b)=(3a+b)(7ab-4)$
если $a=-\frac{1}{3};b=2$:
$(3\ast (-\frac{1}{3})+2)(7\ast (-\frac{1}{3})\ast 2-4)=(-1+2)(-\frac{14}{3}-4)=-4\frac{2}{3}-4=-8\frac{2}{3}$
Ответ: $-8\frac{2}{3}$

$21a^{2}b-4b-12a+7a{b}^2=(21a^{2}b+7a{b}^2)-(12a+4b)=7ab(3a+b)-4(3a+b)=(3a+b)(7ab-4)$
если $a=4;b=\frac{1}{7}$:
$(3a+b)(7ab-4)=(3\ast 4+\frac{1}{7})(7\ast 4\ast \frac{1}{7}-4)=(12+\frac{1}{7})(28\ast \frac{1}{7}-4)=12\frac{1}{7}\ast (4-4)=12\frac{1}{7}\ast 0=0$
Ответ: 0

$21a^{2}b-4b-12a+7a{b}^2=(21a^{2}b+7a{b}^2)-(12a+4b)=7ab(3a+b)-4(3a+b)=(3a+b)(7ab-4)$
если $a=1\frac{1}{7};b=0,5$:
$(3a+b)(7ab-4)=(3\ast 1\frac{1}{7}+0,5)(7\ast 1\frac{1}{7}\ast 0,5-4)=(3\ast \frac{8}{7}+\frac{1}{2})(7\ast \frac{8}{7}\ast \frac{1}{2}-4)=(\frac{24}{7}+\frac{1}{2})(4-4)=(\frac{24}{7}+\frac{1}{2})\ast 0=0$
Ответ: 0

$21a^{2}b-4b-12a+7a{b}^2=(21a^{2}b+7a{b}^2)-(12a+4b)=7ab(3a+b)-4(3a+b)=(3a+b)(7ab-4)$
$a=-\frac{2}{3};b=3$:
$(3\ast (-\frac{2}{3})+3)(7\ast (-\frac{2}{3})\ast 3-4)=(-2+3)(7\ast (-2)-4)=-14-4=-18$
Ответ: −18