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

\(=-2xy+\dfrac{2}{3}x+\dfrac{1}{4}\)

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

\(P=\left(\dfrac{1}{3}x^2y-\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{2}xy^2\right)-\left(xy+5xy\right)=\dfrac{3}{2}xy^2-6xy\)

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Bài giải:

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

\(\Leftrightarrow P=\frac{3}{2}xy^2-6xy\)

Thay \(x=0,5;y=1\)vaof  P; dc:

\(P=\frac{3}{2}\cdot0,5-6.0,5=\frac{1}{2}\left(\frac{3}{2}-\frac{12}{2}\right)=\frac{1}{2}\cdot\frac{-9}{2}=-\frac{9}{4}\)

Đề bài sai rồi bạn

=—11/4xy^2+2x^2y+6/5xy

A=\(\left(3xy^2-2xy^2-4xy^2\right)+\left(2x^2y+\frac{1}{4}x^2y\right)+\left(xy+\frac{1}{5}xy\right)\)

A=\(-3xy^2+\frac{9}{4}x^2y+\frac{6}{5}xy\)

\(P=\dfrac{1}{3}x^2y+xy^2-xy+\dfrac{1}{2}xy^2-5xy-\dfrac{1}{3}x^2y=\dfrac{3}{2}xy^2-6xy\)

Thay x = 2 ; y = 1 ta được 

\(\dfrac{3}{2}.2.1-6.2.1=3-12=-9\)

Thank you..

a/ \(\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

\(=-3xy+4y^2\)

b/ \(\left(x^2-y^2+2xy\right)-\left(x^2+xy+2y^2\right)+\left(4xy-1\right)\)

\(=x^2-y^2+2xy-x^2-xy-2y^2+4xy-1\)

\(=-3y^2+5xy-1\)

a) \(\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

\(=4y^2-3xy\)

b) \(\left(x^2-y^2+2xy\right)-\left(x^2+xy+2y^2\right)+\left(4xy-1\right)\)

\(=x^2-y^2+2xy-x^2-xy-2y^2+4xy-1\)

\(=-3y^2+5xy-1\)

bài này mình giải đc rồi các bạn k cần giải nữa đâu