ta có: \(x^2-y=y^2-x\)\(\Leftrightarrow x^2-y^2=-\left(x-y\right)\)\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=-\left(x-y\right)\)

        \(\Leftrightarrow x+y=-1\)

do đó: \(A=\left(x+y\right)^2-3\left(x+y\right)\)\(=\left(-1\right)^2-3.\left(-1\right)\)\(=4\)

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

\(< =>\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0< =>\left(x-y\right)\left(x+y+1\right)=0\)

\(< =>\orbr{\begin{cases}x-y=0\\x+y+1=0\end{cases}< =>\orbr{\begin{cases}x=y\\x+y=-1\end{cases}}}\)

Mà \(x\ne y=>x+y=-1\)

Vậy \(M=\left(x+y\right)^2-3\left(x+y\right)=\left(-1\right)^2-3.\left(-1\right)=1+3=4\)

Từ \(x^2-y=y^2-x\)\(\Rightarrow x^2-y^2+x-y=0\)

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

\(\Rightarrow\left(x-y\right)\left(x+y+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x+y+1=0\end{matrix}\right.\)\(\Rightarrow x+y=-1\) (vì \(x,y\) là 2 số khác nhau)

Khi đó \(A=x^2+2xy+y^2-3x-3y\)

\(=\left(x+y\right)^2-3\left(x+y\right)=\left(-1\right)^2-3\cdot\left(-1\right)=4\)

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

khi đó:

\(\left[{}\begin{matrix}A=y^2+2y.y+y^2-3y-3y\\A=\left[\left(-1-y\right)+y\right]^2-3\left(-1-y+y\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}A=4y^2-6y\\A=4\end{matrix}\right.\)

đến đây thì mình chả bt trình bày sao nửa, mong bạn thông cảm

Ta co:\(x^2-y=y^2-x\)

\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\x+y=-1\end{cases}}\)

TH:\(x=y\left(l\right)\)(Vi x,y la 2 so khac nhau)

TH:\(x+y=-1\)

Ta co:\(A=\left(x+y\right)^2-3\left(x+y\right)=1+3=4\)

M= x²  + 2xy - 3x - 3y + y² + 10

= x² + 2xy + y² - 3(x + y) + 10

= (x + y)² - 3(x + y) + 10

= 4 - 6 + 10 ( do x + y = 2 )

= 8

Vậy M = 8