Đặt A=\(\left(-2\dfrac{1}{3}\cdot x^2y^3\right)\left(\dfrac{9}{14}xy^2\right)\)

\(=-\dfrac{7}{3}\cdot x^2y^3\cdot\dfrac{9}{14}\cdot xy^2\)

\(=\left(-\dfrac{7}{3}\cdot\dfrac{9}{14}\right)\cdot x^2\cdot x\cdot y^3\cdot y^2=\dfrac{-3}{2}x^3y^5\)

Khi x=1 và y=-1 thì \(A=\dfrac{-3}{2}\cdot1^3\cdot\left(-1\right)^5=-\dfrac{3}{2}\cdot1\cdot\left(-1\right)=\dfrac{3}{2}\)

\(\left(-\dfrac{1}{4}x^2y^3\right)\left(\dfrac{6}{5}xy^2\right)\\ =\left(-\dfrac{1}{4}.\dfrac{6}{5}\right)\left(x^2.x\right)\left(y^3.y^2\right)\\ =-\dfrac{3}{10}x^3y^5\)

Hệ số: \(-\dfrac{3}{10}\)

Biến:x³y⁵

Bậc:8

Ta có: \(3x^2\cdot\left(\dfrac{1}{6}\right)^2\cdot y^2\)

\(=3\cdot\dfrac{1}{36}\cdot x^2\cdot y^2\)

\(=\dfrac{1}{12}x^2y^2\)

Hệ số là \(\dfrac{1}{12}\)

Phần biến là \(x^2;y^2\)

Bậc là 4

Bài 7 

\(-3y\left(x^2y^2\right)\left(-x^3y^9\right)=3x^5y^{12}\)

hệ sô : 3 ; biến x^5y^12 ; bậc 17 

lx