Đã làm được, thank các bác nhiều!

Lời giải:

$2Q=2x^2+2xy+2y^2-6x-6y+3998$

$=(x^2+2xy+y^2)+x^2+y^2-6x-6y+3998$

$=(x+y)^2-4(x+y)+(x^2-2x)+(y^2-2y)+3998$

$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+3992$

$=(x+y-2)^2+(x-1)^2+(y-1)^2+3992\geq 3992$

$\Rightarrow Q\geq 1996$

Vậy $Q_{\min}=1996$ khi $x+y-2=x-1=y-1=0\Leftrightarrow x=y=1$

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$R=(x^2+2xy+y^2)+x^2-2x+2y+15$

$=(x+y)^2+2(x+y)+x^2-4x+15$

$=(x+y)^2+2(x+y)+1+(x^2-4x+4)+10$

$=(x+y+1)^2+(x-2)^2+10\geq 10$
Vậy $R_{\min}=10$ khi $x+y+1=x-2=0$

$\Leftrightarrow x=2; y=-3$

Max:

\(M=\frac{x^2+xy+y^2}{x^2+y^2}=1+\frac{xy}{x^2+y^2}\le1+\frac{xy}{2\left|xy\right|}\le1+\frac{xy}{2xy}=1+\frac{1}{2}=\frac{3}{2}\)

Dấu "=" xảy ra tại x=y

* \(3x+y=1\Rightarrow y=1-3x\)

\(M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=12\left(x^2-\dfrac{1}{2}x+\dfrac{1}{12}\right)=12\left(x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}\right)+\dfrac{1}{4}=12\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)\(\Rightarrow Min_M=\dfrac{1}{4}\Leftrightarrow x=y=\dfrac{1}{4}\)

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

\(4N=4x^2+4xy+4y^2-12x-12y\)

\(4N=\left(4x^2+4xy+y^2\right)-12x-6y+9+3y^2-6y+3-12\)

\(4N=\left(2x+y\right)^2-2.3\left(2x+y\right)+9+3\left(y-1\right)^2-12\)

\(4N=\left(2x+y-3\right)^2+3\left(y-1\right)^2-10\ge-12\)

\(\Rightarrow N\ge-3\)

\(\Rightarrow Min_N=-3\Leftrightarrow x=y=1\)

\(M=5xy+\dfrac{3}{2}x^2-6y^2+2xy-4y^2+3x^2\)

\(M=\left(5xy+2xy\right)+\left(\dfrac{3}{2}x^2+3x^2\right)+\left(-6y^2-4y^2\right)\)

\(M=7xy+\dfrac{9}{2}y^2-10x^2\)