Đáp án: a.Bậc $10$, Hệ số $-\dfrac1{32}a^4b^2$
b.Bậc $8$, Hệ số $-\dfrac{64a}{5b}$
Giải thích các bước giải:
a.$M=(-2axyx^2)\cdot(4axy)(\dfrac{-1}{16}abx^2)^2$
$\to M=-2ax^{1+2}y\cdot 4axy\cdot (\dfrac{-1}{16})^2a^2b^2x^4$
$\to M=-2ax^{3}y\cdot 4axy\cdot \dfrac{1}{256}a^2b^2x^4$
$\to M=(-2\cdot 4\cdot \dfrac{1}{256})\cdot (a\cdot a\cdot a^2)\cdot b^2\cdot (x^3\cdot x\cdot x^4)\cdot (y\cdot y)$
$\to M=-\dfrac1{32}\cdot a^{1+1+2}\cdot b^2\cdot \cdot x^{3+1+4}\cdot y^{1+1}$
$\to M=-\dfrac1{32}\cdot a^4\cdot b^2\cdot x^{8}\cdot y^{2}$
$\to$Hệ số $-\dfrac1{32}a^4b^2$
Bậc $8+2=10$
b.Ta có :
$N=\dfrac{16\cdot (-2x^2y)^2\cdot (-axy)}{5b}$
$\to N=\dfrac{16\cdot 4x^4y^2\cdot (-a)xy}{5b}$
$\to N=\dfrac{-64a\cdot 4x^{4+1}y^{2+1}}{5b}$
$\to N=\dfrac{-64a\cdot 4x^{5}y^{3}}{5b}$
$\to $Bậc $5+3=8$
Hệ số $-\dfrac{64a}{5b}$
