\(\dfrac{2}{5}x^2y+xy^2-3xy+\dfrac{1}{3}xy^2-3xy-\dfrac{1}{2}x^2y\)

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

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

`(xy-5)(xy+2)+3(xy-2)(xy+2)-(3xy-1/2)^2 +5x^2y^2`

`=(xy+2)(xy-5+3xy-6)-(9x^2y^2-3xy+1/4)+5x^2y^2`

`=(xy+2)(4xy-11)-9x^2y^2+3xy-1/4+5x^2y^2`

`=4x^2y^2-11xy+8xy-22-4x^2y^2+3xy-1/4`

`=-89/4`

Ta có: \(\left(xy-5\right)\left(xy+2\right)+3\left(xy-2\right)\left(xy+2\right)-\left(3xy-\dfrac{1}{2}\right)^2+5x^2y^2\)

\(=x^2y^2-3xy-10+3x^2y^2-12-\left(9x^2y^2-3xy+\dfrac{1}{4}\right)+5x^2y^2\)

\(=9x^2y^2-3xy-22-9x^2y^2+3xy-\dfrac{1}{4}\)

\(=\dfrac{-89}{4}\)

bài này quá dễ nên mik không thick làm

nhớ giải giùm mình nha 

hộ cái đi mà

N = -1/10 x² y +4/3xy² -6xy

đúng 100000000000000000000000000% luôn bạn nhé

#)Giải :

a) \(M=\left(3x^3+3x^2y-3xy^2+xy\right)-\left(2x^3-3x^2y-3xy^2+xy+1\right)\)

\(M=\left(3x^3-2x^3\right)+\left(3x^2y-3x^2y\right)+\left(-3xy^2+3xy^2\right)-\left(xy-xy\right)+1\)

\(M=x^3+1\)

b) \(M=-28\Leftrightarrow1+x^3=-28\)

\(\Rightarrow x^3=-27=\left(-3\right)^3=-3\)

Vậy ..................................................

trả lời 

B=-3

chúc bn 

hc tốt

trả lời hộ với mai thi rồi

mũ 4

\(\dfrac{1}{2}xy.3xy^2\\ =\dfrac{3}{2}x^2y^3\)

\(\dfrac{1}{2}xy\text{×}3xy^2=\dfrac{3}{2}x^2y^3\)

C

C

Chọn C