Có vẻ đề  đúng

\(P=\frac{3x^2y-1}{4xy}\)

\(\left(x^2+y^2+1^2-2xy-2x+2y\right)+\left(y^2+4y+4\right)=0\)

\(\left(x+y-1\right)^2+\left(y+2\right)^2=0\)

\(\hept{\begin{cases}x+y-1=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=-2\end{cases}\Rightarrow}P=\frac{3.9.\left(-2\right)-1}{4.3.\left(-2\right)}=\frac{55}{24}}\)

Cách giải đúng rồi nhưng sai hằng đảng thức nha bạn 
\(x^2+y^2+1-2xy-2x+2y=\left(y-x+1\right)^2\)

rồi sửa x= -1 là được

đề bài bạn sai 

\(gt\Leftrightarrow\left(x-y-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow x=-1;y=-2\)

Done !!

(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 :

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

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

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

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

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

\(2xy+2x-5z=0\Leftrightarrow z=\frac{2xy+2x}{5}\)

Sau đấy bn thay z vào là ra 

Ta có: \(2xy+2x-5z=0\Rightarrow z=\frac{2xy+2x}{5}\)

Thay \(z=\frac{2xy+2x}{5}\)vào A, ta được: \(A=x^2+2y^2+2xy+\frac{8}{5}y+\frac{2xy+2x}{5}+2=x^2+2y^2+\frac{12}{5}xy+\frac{8}{5}y+\frac{2}{5}x+2\)\(=\left(x^2+\frac{12}{5}xy+\frac{36}{25}y^2\right)+\frac{2}{5}\left(x+\frac{6}{5}y\right)+\frac{1}{25}+\left(\frac{14}{25}y^2+\frac{28}{25}y+\frac{14}{25}\right)+\frac{7}{5}\)\(=\left[\left(x+\frac{6}{5}y\right)^2+\frac{2}{5}\left(x+\frac{6}{5}y\right)+\frac{1}{25}\right]+\frac{14}{25}\left(y+1\right)^2+\frac{7}{5}\)\(=\left(x+\frac{6}{5}y+\frac{1}{5}\right)^2+\frac{14}{25}\left(y+1\right)^2+\frac{7}{5}\ge\frac{7}{5}\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x+\frac{6}{5}y+\frac{1}{5}=0\\y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\Rightarrow z=0\)

\(2x^2+2y^2+z^2-2x+2y+2xy+2yz+2zx+2=0\)

\(\Leftrightarrow\)\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+y\right)^2+\left(y+z\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\)\(x=-y=z=1\)

\(\Rightarrow\)\(A=x^{2018}+y^{2018}+z^{2018}=1^{2018}+\left(-1\right)^{2018}+1^{2018}=3\)

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