Question

cho x, y la 2 số khác nhau thoã mãn x^2 -y = y^2-x. Tính giá trị biểu thức

A= x^3+y^3-3xy(x^2+y^2)+6x^2y^2(x+y)

Answer 1

\(x^2-y=y^2-x\Leftrightarrow x^2-y^2+x-y=0\Leftrightarrow\left[{}\begin{matrix}x=y\left(l\right)\\x+y+1=0\left(nh\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=\left(x+y\right)\left(x^2-xy+y^2\right)-3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=-\left(x^2-xy+y^2\right)-3xy\left(x^2+y^2\right)-6x^2y^2\\x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=-\left(1-3xy\right)-3xy\left(1-2xy\right)-6x^2y^2\\x^2+y^2=1-2xy\end{matrix}\right.\)

\(\Leftrightarrow A=-1+3xy-3xy+6x^2y^2-6x^2y^2=-1\)