=-1/2x^2+5x^2y^3-8x^3y^2-5x^2y^3+7x^3y^2-6x^2-5/3y

=(-1/2x^2+6x^2)+(5x^2y^3-5x^2y^3)+(-8x^3y^2-7x^3y^2)+5/3y

=11/2x^2+0-15x^3y^2+5/3y

=11/2x^2-15x^3y^2+5/3y

thay x=-1/2 , y=25 vào giá trị biểu thức M ta đc

       11/2.(-1/2)^2-15.(-1/2)^3.25^2+5/3.25=7273/6

   vậy tại x=-1/2 , y=25 vào giá trị biểu thức M có giá trị là 7273/6

Em kiểm tra lại đề bài nhé vì:

\(Q=\left(x^3.x.y^n.y-\frac{1}{2}x^3.y^n.y^2\right):\frac{1}{2}x^3y^n-\left(4.5.x^2.x^2.y\right):\left(5x^2y\right)\)

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

\(=2xy-y^2-4x^2=-\left(x^2-2xy+y^2\right)-3x^2=-\left[\left(x-y\right)^2+3x^2\right]< 0\)Với mọi x, y khác 0

=> Q luôn có gia trị âm với mọi x, y khác 0.

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

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

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

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

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

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

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 

\(\frac{5x-2y}{x+3y}=\frac{7}{4}\)

=> (5x - 2y).4 = 7.(x + 3y)

=> 20x - 8y = 7x + 21y

=>> 20x - 7x = 21y + 8y

=> 13x = 29y

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

\(\frac{5x-2y}{x+3y}=\frac{7}{4}\)

\(\Rightarrow4\left(5x-2y\right)=7\left(x+3y\right)\)

\(\Rightarrow20x-8y=7x+21y\)

\(\Rightarrow20x-7x=8y+21y\)

\(\Rightarrow13x=29y\)

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

Vậy \(\frac{x}{y}=\frac{29}{13}\)

\(\frac{5x-2y}{x+3y}=\frac{7}{4}\)

\(\Rightarrow7\left(x+3y\right)=4\left(5x-2y\right)\)

\(\Rightarrow7x+21y=20x-8y\)

\(\Rightarrow7x-20x=-8y-21y\)

\(\Rightarrow-13x=-29y\)

\(\Rightarrow13x=29y\)

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

Vậy : \(\frac{x}{y}=\frac{29}{13}\)

Từ đề bài \(\Rightarrow\)\(x^2-2y^2-xy=0\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)

Mà \(x+y\ne0\Rightarrow x-2y=0\Rightarrow x=2y\)

\(\Rightarrow P=\frac{2y-y}{2y+y}=\frac{1}{3}\)

Vì \(x^2-2y^2=xy\) 

\(\Leftrightarrow x^2-xy-y^2=0\)

\(\Leftrightarrow\left(x-y\right)^2-y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)

Theo đề bài thì có : 

\(x+y\ne0\)

\(\Rightarrow x-2y=0\)

\(\Leftrightarrow x=2y\)

Từ đó ta lại có :

\(P=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

Vậy .......

ta có 

          x²-2y²=xy

<=>  x² -xy -2y² =0

<=> (x-2y)(x+y)=0

=>\(\orbr{\begin{cases}x=2y\\x+y=0\left(loại\right)\end{cases}}\)

nếu x=2y thì P=1/3

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

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

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

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

\(=-\left(x+1\right)+\frac{x^2}{x+1}\)`

\(=\frac{-\left(x+1\right)^2+x^2}{x+1}\)

\(=\frac{-x^2-2x-1+x^2}{x+1}\)

\(=\frac{-2x-1}{x+1}\)(1)

b) Thay \(x=-3,y=2014\)vào (1) ta được:

\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)

Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014