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(x + y - 5)² + 2(y - 1)² - 9 = 0

<=> 2(y - 1)² = 9 - (S - 5)² \(\ge0\)

\(\Leftrightarrow\left(S-5\right)^2\le9\)

\(\Leftrightarrow-3\le S-5\le3\)

\(\Leftrightarrow2\le S\le8\)

Vậy GTNN là 2 đạt được khi x = y = 1

GTLN là 8 đạt được khi (x, y) = (7, 1)

\(x^2+3y^2+2xy-10x-14y+18\)

\(\Rightarrow\left(x^2+2xy-10x+y^2-10y+25\right)+2y^2-4y-7=0\)

\(\Rightarrow\left(x+y-5\right)^2+2y^2-4y+2-9=0\)

\(\Rightarrow\left(x+y-5\right)^2+2\left(y^2-2y+1\right)-9=0\)

\(\Rightarrow\left(x+y-5\right)^2+2\left(y-1\right)^2-9=0\)

....

x=7;y=±1 và x=y=1 và x=1; y=3 và x=y=3 và x=5;y=-1

Ta có : 

\(x^2+3y^2+2xy-10x-14y+18=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)-10x-10y+25+\left(2y^2-4y+2\right)-9=0\)

\(\Leftrightarrow\left(x+y\right)^2-2.\left(x+y\right).5+25+2\left(y^2-2y+1\right)=9\)

\(\Leftrightarrow\left(x+y-5\right)^2+2\left(y-1\right)^2=9\)

Vì \(2\left(y-1\right)^2\ge0\forall y\)nên  \(\left(x+y-5\right)^2\le9\)hay \(\left(M-5\right)^2\le9\)

\(\Rightarrow-3\le M-5\le3\Leftrightarrow2\le M\le8\)

\(E=1983-x^2-3y^2+2xy-10x+14y\)

\(-E=x^2+3y^2-2xy+10x-14y-1983\)

\(-E=\left(x^2-2xy+y^2\right)+2y^2+10x-14y-1983\)

\(-E=\left[\left(x-y\right)^2+2\left(x-y\right).5+25\right]\)\(+2\left(y^2-2y+1\right)+1956\)

\(-E=\left(x-y+5\right)^2+2\left(y-1\right)^2+1956\)

Do  \(\left(x-y+5\right)^2\ge0\forall x;y\)

             \(2\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow-E\ge1956\Leftrightarrow E\le-1956\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x-y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=1\end{cases}}\)

Vậy ...

\(x^2-3y^2-2xy+10x+14y-18\)

\(=x^2-2xy+y^2-2x^2+10x-4y^2+14y-18\)

\(=x^2-2xy+y^2-2\left(x^2-5x+25\right)-4\left(y^2-\frac{7}{2}y+\frac{49}{16}\right)+\frac{177}{4}\)

\(=\left(x-y\right)^2-2\left(x-5\right)^2-4\left(y-\frac{7}{4}\right)^2+\frac{177}{4}\)

.....

-2A=2x²+6y²+4xy-20x-28y+36

=(x²+4xy+4y²)+(x²-20x+100)+2(y²-14y+49)-162

=(x+2y)²+(x-10)²+2(y-7)²-162\(\ge\)-162

=> A\(\le81\)

Dấu "=" xảy ra khi

Giups mk vs ạ ai nhanh mk tick nha

Lời giải:
\(x^2+3y^2+10x-14y-2xy=11\)

$\Leftrightarrow (x^2-2xy+y^2)+2y^2+10x-14y=11$

$\Leftrightarrow (x-y)^2+10(x-y)+25+(2y^2-4y+2)=38$

$\Leftrightarrow (x-y+5)^2+2(y-1)^2=38$

$\Rightarrow (x-y+5)^2=38-2(y-1)^2\leq 38$

$\Rightarrow -\sqrt{38}\leq x-y+5\leq \sqrt{38}$

$\Leftrightarrow -\sqrt{38}-5\leq x-y\leq \sqrt{38}-5$
Vậy $A_{\min}=-\sqrt{38}-5$ và $A_{\max}=\sqrt{38}-5$