Question

[Đỗ thị như quỳnh]

Viet moi don thuc sau thanh don thuc rut gon , cho ro phan bien va he so

a, \(2x^2y^3.\dfrac{1}{4}xy^3\left(-3xy\right)\)

b, \(\left(-2x^3y\right)^2.xy^2.\dfrac{1}{5}y^5\)

Answer 1

[Hoàng Thị Ngọc Anh]

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

\(=\left(-3.2.\dfrac{1}{4}\right)x^4y^7\)

\(=\dfrac{-3}{2}x^4y^7\)

\(\Rightarrow Hệ\) số: \(\dfrac{-3}{2}\)

Phần biến: \(x^4y^7\)

b) \(\left(-2x^3y\right)^2.xy^2.\dfrac{1}{5}y^5\)

\(=\dfrac{4}{5}x^7y^9\)

\(\Rightarrow Phần\) biến: \(x^7y^9\)

Hệ số: \(\dfrac{4}{5}.\)

Answer 2

[Aki Tsuki]

a/ \(2x^2y^3\cdot\dfrac{1}{4}xy^3\left(-3xy\right)\)

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

\(=-\dfrac{3}{2}x^4y^7\)

Phần biến: \(x^4y^7\)

Hệ số: \(-\dfrac{3}{2}\)

b/ \(\left(-2x^3y\right)^2\cdot xy^2\cdot\dfrac{1}{5}y^5=4x^6y^2\cdot xy^2\cdot\dfrac{1}{5}y^5\) \(=4\cdot\dfrac{1}{5}\left(x^6\cdot x\right)\left(y^2\cdot y^2\cdot y^5\right)=\dfrac{4}{5}x^7y^9\)

Phần biến: \(\dfrac{4}{5}\)

Hệ số: \(x^7y^9\)