Question

(4a^3*b^4)^2*(8a^2*b^3)^3/(16a^4b^4)^2 , якщо a= -1/2, b= -1

Answer 1

$frac{(4a^{3}b^{4})^{2} *(8a^{2}b^{3})^{3} }{(16a^{4}b^{4})^{2}} = frac{2^{4} a^{6}b ^{8}*8^{3}a^{6}b^{9}}{2^{2}*8^{2}a^{8}b^{8}}= frac{2^{4}*8^{3}a^{12}b^{17}}{2^{2}*8^{2}a^{8}b^{8}}= 32a^{4}b^{9}$

при  а= - 1/2,  b= - 1
$32*( frac{1}{2})^{4} *( -1)^{9}= - frac{32}{16} = -2$