Đặt \(xy-12x+15y\)là (*)

Từ phương trình (1) ta có \(x^2-3xy+2y^2+x-y=0\Leftrightarrow\left(x-y\right)\left(x-2y\right)+\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x-2y+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=y\\x=2y-1\end{cases}}\)

Với \(x=y\)thay vào (2) ta có \(x^2-2x^2+x^2-5x+7x=0\Leftrightarrow x=0\Rightarrow x=y=0\)

Thay \(x=y=0\)vào (*) ta thấy 0.0-12.0+15.0=0(tm)

Với \(x=2y-1\Rightarrow\left(2y-1\right)^2-2\left(2y-1\right)y+y^2-5\left(2y-1\right)+7y=0\)

\(\Leftrightarrow4y^2-4y+1-4y^2+2y+y^2-10y+5+7y=0\)

\(\Leftrightarrow y^2-5y+6=0\Leftrightarrow\left(y-2\right)\left(y-3\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}}\)

Với \(x=3;y=2\)thay vào (*)  ta thấy \(3.2-12.3+15.0=0\left(tm\right)\)

Với \(x=5;y=3\)thay vào (*)  ta thấy \(5.3-12.5+15.3=0\left(tm\right)\)

Vậy .....

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Hơn 170 điểm đã  là CTV rồi 

Áp dụng BĐT Cô-si:

\(x^3+x^3+8\ge3\sqrt[3]{8x^6}=6x^2\)

\(y^6+y^6+1+1+1+1\ge6\sqrt[6]{y^{12}}=6y^2\)

Cộng vế:

\(2\left(x^3+y^6\right)+12\ge6\left(x^2+y^2\right)\ge30\)

\(\Rightarrow x^3+y^6\ge9\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;1\right)\)

\(xy-\left(x+2y\right)=3\)
\(xy-x-2y=3\)
\(y\left(x-2\right)-x=3\)
\(y\left(x-2\right)-x+2=3+2\)
\(y\left(x-2\right)-\left(x-2\right)=5\)
\(\left(y-1\right)\left(x-2\right)=5\)
Ta có bảng sau:

Vậy các cặp \(\left(x;y\right)\) là \(\left(7;2\right);\left(3;6\right);\left(-3;0\right);\left(1;-4\right)\)

=>xy-x-2y=3

=>x(y-1)-2y+2=5

=>(x-2)(y-1)=5

=>\(\left(x-2;y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;6\right);\left(7;3\right);\left(1;-4\right);\left(-3;0\right)\right\}\)

Tá có : $(x+1).(y+1).(z+1) = (x-1).(y-1).(z-1)$

$\to xyz+1+x+y+z+xy+yz+zx =xyz + x + y + z -xy-yz-zx-1$

$\to 2.(xy+yz+zx) = -2$

$\to xy+yz+zx=-1$

\(2x^2+2y^2\ge4xy\)

\(4x^2+z^2\ge4xz\)

\(4y^2+z^2\ge4yz\)

Cộng vế:

\(2\left(3x^2+3y^2+z^2\right)\ge4\left(xy+yz+zx\right)\ge20\)

\(\Rightarrow3x^2+3y^2+z^2\ge10\)

Dấu "=" xảy ra tại \(\left(x;y;z\right)=\left(1;1;2\right);\left(-1;-1;-2\right)\)