\(x^2+2y^2-3xy=0\Leftrightarrow\left(x-y\right)\left(x-2y\right)=0\)

\(\Leftrightarrow x-2y=0\) (do \(x>y\) nên \(x-y>0\))

\(\Leftrightarrow x=2y\)

\(\Rightarrow A=\dfrac{6.2y+16y}{5.2y-3y}=\dfrac{28y}{7y}=4\)

\(x^2+2y^2-3xy=0\)

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

\(\Rightarrow\left(x-2y\right)\left(x-y\right)=0\Rightarrow\orbr{\begin{cases}x=2y\\x=y\end{cases}}\)

x = 2y thì \(A=\frac{2018.2y.y}{\left(2y\right)^2+2y^2}=\frac{4036y^2}{6y^2}=\frac{2018}{3}\)

x = y thì \(A=\frac{2018.y.y}{y^2+y^2}=\frac{2018y^2}{2y^2}=1009\)

Vậy \(\orbr{\begin{cases}A=\frac{2018}{3}\\A=1009\end{cases}}\)

Ta có: x²+y=y²+x

=>x²+y-y²+x=0

=>(x²-y²)-(x-y)=0

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

=>(x-y)(x+y-1)=0

=>x-y=0 hoặc x+y-1=0

=>x+y=1(TH1 loại do x khác y)

ta có:A=x³+y³+3xy(x²+y²)+6x²y²(x+y)

=>A=(x+y)(x²-xy+y²)+3x³y+3xy³+6x²y²

=>A=x²-xy+y²+3x³y+3xy³+6x²y²

=>A=(x+y)²-3xy+3x²y(x+y)+3xy²(x+y)

=>A=1-3xy+3x²y+3xy²

^{=>A=1+3xy(-1+a+b)}

^{=>A=1+3xy(-1+1)}

^=>A=1+3xy.0

^=>A=1

Vậy A=1 khi x²+y=y²+x và x khác y.

Lê Đức Huy chép sai đề cau đầu kìa!

(x² +y² +9+2xy-6x-6y)+(y²+4y+4)=0

(x+y-3)²+(y+2)²=0.vì (x+y-3)²>=0;(y+2)²>=0

suy ra x+y-3=0 và y+2=0

x=5;y=-2

thay x,y vào bt H ta đc H=1

x² + 2y² + 2xy - 6x - 2y + 13 = 0

<=> ( x² + 2xy + y² - 6x - 6y + 9 ) + ( y² + 4y + 4 ) = 0

<=> [ ( x² + 2xy + y² ) - ( 6x + 6y ) + 9 ] + ( y + 2 )² = 0

<=> [ ( x + y )² - 2( x + y ).3 + 3² ] + ( y + 2 )² = 0

<=> ( x + y - 3 )² + ( y + 2 )² = 0

Ta có : \(\hept{\begin{cases}\left(x+y-3\right)^2\\\left(y+2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x+y-3\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra <=> x = 5 ; y = -2

Thế x = 5 ; y = -2 vào A ta được :

\(A=\frac{5^2-7\cdot5\cdot\left(-2\right)+52}{5-\left(-2\right)}=\frac{25+70+52}{7}=\frac{147}{7}=21\)