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

P=0+3/2xy^2+6xy 

P=3/2xy^2+6xy

Thu gọn rồi tính giá trị của đa thức P tại x = 0,5 và y = 1.

Ta có: P = x² y + xy² – xy + xy² – 5xy – x²y

P = x² y – x²y + xy² + xy² – xy – 5xy = xy² – 6xy

Thay x = 0,5 và y = 1 ta được

P = . 0,5 . 1² – 6. 0,5 . 1 = - 3 = .

Vậy P = tại x = 0,5 và y = 1.

đầu tiên ta thu gọn trước để tính cho dễ:

 Ta có:           

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

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

        \(P=\frac{3}{2}xy^2-6xy\)

rồi thay x = 0,5 và y = 1 vào

   ta đc: 

              \(P=\frac{3}{2}\cdot\left(0,5\right)\cdot1^2-6\cdot\left(0,5\right)\cdot1=0,75-3=-2,25\)

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

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

\(=0+xy^2\left(1+\frac{1}{2}\right)-xy\left(1+5\right)\)

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

Thay x = 0,5 và y = 1 vào đa thức trên ta có:

\(\frac{3}{2}.0,5.1^2-6.0,5.1\)

\(=\frac{3}{2}.\frac{1}{2}.1-6.\frac{1}{2}.1\)

\(=\frac{3}{4}-\frac{6}{2}=\frac{3}{4}-\frac{12}{4}=-\frac{9}{4}\)

P/s: Ko chắc!

P=\(\frac{1}{3}\)\(^{x^2}\)y+\(^{xy^2}\)−xy+\(\frac{1}{2}\)\(^{xy^2}\)−5xy−\(\frac{1}{2}\)\(^{xy^2}\)

P=(\(\frac{1}{3}\)−\(\frac{1}{3}\))+(\(^{xy^2}\)+\(\frac{1}{2}\)\(^{xy^2}\))+(−xy−5xy)

P=0+\(\frac{3}{2}\)\(^{xy^2}\)+(−6xy)=\(\frac{3}{2}\)\(^{xy^2}\)−6xy

Thay x=0,5;y=1 vào P ta có:

\(\frac{3}{2}\).0,5.\(^{1^2}\)−6.0,5.1=\(\frac{3}{4}\)−3=\(\frac{3}{4}\)−\(\frac{12}{4}\)=\(\frac{-9}{4}\)=−2,25

Vậy P=\(\frac{-9}{4}\)=-2,25

\(p=\frac{1}{3}x^2y+xy^2-xy+\frac{1}{2}xy^2-5xy-\frac{1}{3}x^2y\)

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

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

thay x = 0,5 và y = 1 vào P

\(\Rightarrow\)\(=\frac{3}{2}.0,5.1^2-6.0,5.1\)

\(=\frac{3}{2}.0,5-6.0,5\)

\(=\left(\frac{3}{2}-6\right).0,5\)

\(=\frac{-9}{2}.0,5\)

\(=\frac{-9}{4}\)

~hok tốt ~

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

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

\(=\frac{11}{2}x^2y+xy+\frac{1}{6}x+\frac{1}{2}\)

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

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

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

\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)

\(=\frac{\left(x+y\right)}{y}-\frac{x^2}{y\left(x+y\right)}\)

\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)

Giờ chỉ cần thế x, y vô nữa là xong nhé.

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

\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)

\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)

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

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

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

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

\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)

\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)

\(=\frac{x+y}{y}-\frac{x^2}{y\left(x+y\right)}=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}\)

\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)

\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)

Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :

\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)

Ừ nhở chị sai từ chỗ \(\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{x^2+2xy+y^2-x^2}{y\left(x+y\right)}=\frac{y^2+2xy}{y\left(x+y\right)}\)em nhé