\(A=\left(-\dfrac{2}{3}x^3y^4\right)^2.\left(-3x^5y^2\right)^3\)

\(A=\left(\dfrac{4}{9}x^6y^8\right).\left(-27x^{15}y^6\right)\)

\(A=\left(\dfrac{4}{9}.-27\right)\left(x^6.x^{15}\right)\left(y^8.y^{16}\right)\)

\(A=-12x^{21}y^{24}\)

\(\text{Hệ số:-12}\)

\(\text{Bậc:45}\)

\(B=\left(3x^2y\right).\left(-\dfrac{1}{3}x^3y\right).\left(-\dfrac{1}{4}x^3y^4\right)\)

\(B=\left(3.-\dfrac{1}{3}.-\dfrac{1}{4}\right).\left(x^2.x^3.x^3\right).\left(y.y.y^4\right)\)

\(B=\dfrac{1}{4}x^8y^6\)

\(\text{Hệ số:}\dfrac{1}{4}\)

\(\text{Bậc:14}\)

a) \(A=\left(-\dfrac{1}{4}ax^3\right)\cdot\left(-8xy^2\right)^0\)

\(=-\dfrac{1}{4}a\cdot x^3\)

Hệ số là \(-\dfrac{1}{4}a\)

Bậc là 3

a) Ta có: \(A=1\dfrac{1}{4}\cdot x^3y\cdot\left(-\dfrac{6}{7}xy^5\right)^0\cdot\left(-2\dfrac{2}{3}xy\right)\)

\(=\dfrac{5}{4}x^3y\cdot\dfrac{-8}{3}xy\)

\(=\left(\dfrac{5}{4}\cdot\dfrac{-8}{3}\right)\cdot\left(x^3\cdot x\right)\cdot\left(y\cdot y\right)\)

\(=\dfrac{-10}{3}x^4y^2\)

a, (-3a⁶)(-1/2a³)x⁸y⁶

-Hệ số: (-3a⁶)(-1/2a³)

-Biến: x⁸y⁶

b, -Bậc: 14.

a: \(A=\dfrac{-1}{3}\cdot\dfrac{-2}{3}\cdot\dfrac{3}{2}\cdot x^2y\cdot xy^3\cdot xy^2=\dfrac{1}{3}x^4y^6\)

Hệ số là 1/3

Bậc là 10

b: \(P=\dfrac{1}{25}+3\cdot\dfrac{1}{5}\cdot\left(-1\right)+1=\dfrac{26}{25}-\dfrac{3}{5}=\dfrac{26}{25}-\dfrac{15}{25}=\dfrac{11}{25}\)

a)A= -\(\dfrac{1}{3}\)x²y. -\(\dfrac{2}{3}\)xy³. 1\(\dfrac{1}{2}\)xy²

A=(-\(\dfrac{1}{3}\).-\(\dfrac{2}{3}\).1\(\dfrac{1}{2}\)). (x².x.x). (y.y³.y²)
A=1.x⁴.y⁶

^{_Hệ số là:1
_Bậc của đơn thức A là 10}
 

b)Thay x=\(\dfrac{1}{5}\) , y=-1vào biểu thức P, ta có:

P=\(\dfrac{1}{5}\).2+3.\(\dfrac{1}{5}\).-1+-1.2
P=\(\dfrac{2}{5}\)+ -\(\dfrac{3}{5}\) +-2
P=-\(\dfrac{1}{5}\)+ -2=-\(\dfrac{11}{5}\)

a: A=-2/3x^4y^3

Hệ số: -2/3

Bậc: 7

b: Khi x=-1 và y=1 thì A=-2/3

\(A=\dfrac{2}{5}x^7y^3\)

Hệ số: \(\dfrac{2}{5}\)

Bậc: 10