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

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

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

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

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

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 ...

gợi ý nhé:

[-(x-y)²-10(x-y)-25] - 2(y-1)² + 2010

= -[(x-y)+5]²  - 2(y-1)² + 2010

tự cậu suy ra MAX nhé

chưa hiểu thì hỏi nhé

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

\(C=-\left(x^2+3y^2-2xy+10x-14y-1983\right)\)

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

\(C=-\left[\left(x-y\right)^2+2\cdot\left(x-y\right)\cdot5+25+2y^2-4y+2-2010\right]\)

\(C=-\left[\left(x-y+5\right)^2+2\left(y-1\right)^2-2010\right]\)

\(C=2010-\left[\left(x-y+5\right)^2+2\left(y-1\right)^2\right]\le2010\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\)

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 $A_{\max}=\sqrt{38}-5$

1.

Đặt \(x+y=a\Rightarrow y=a-x\)

\(\Rightarrow x^2+2x\left(a-x\right)-14\left(a-x\right)-10x+3\left(a-x\right)^2+27=0\)

\(\Leftrightarrow2x^2-4\left(a+1\right)x+3a^2-10a+27=0\)

\(\Delta'=4\left(a+1\right)^2-2\left(3a^2-10a+27\right)\ge0\)

\(\Leftrightarrow-a^2+14a-25\ge0\)

\(\Rightarrow7-2\sqrt{6}\le a\le7+2\sqrt{6}\)

\(\Rightarrow-10-2\sqrt{6}\le P\le-10+2\sqrt{6}\)

2. Chắc đề là \(a;b>0\) (đảm bảo mẫu dương) chứ ko phải \(a.b>4\)

\(M\ge\dfrac{\left(a+b\right)^2}{a+b-8}=\dfrac{\left(a+b-8+8\right)^2}{a+b-8}=\dfrac{\left(a+b-8\right)^2+16\left(a+b-8\right)+64}{a+b-8}\)

\(M\ge a+b-8+\dfrac{64}{a+b-8}+16\ge2\sqrt{\dfrac{64\left(a+b-8\right)}{a+b-8}}+16=32\)

Dấu "=" xảy ra khi \(a=b=8\)

Ta có:

\(A=1993-x^2-3y^2+2xy-10x+14y\\ =2020-\left(x^2-2xy+y^2\right)-10\left(x-y\right)-25-\left(2y^2-4y+2\right)\\ =2020-\left(x-y-5\right)^2-2\left(y-1\right)^2\)

Với mọi x; y thì \(2020-\left(x-y-5\right)^2-2\left(y-1\right)^2\ge2020\)

Để A=2020 thì

\(\left\{{}\begin{matrix}x-y=5\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=1\end{matrix}\right.\)

Vậy...

Đặt  \(A=-x^2-3y^2-2xy+10x+14y-18\)

Ta có : \(-A=x^2+3y^2+2xy-10x-14y+18\)

\(-A=\left(x^2+2xy+y^2\right)+2y^2-10x-14y+18\)

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)\times5+25\right]+2y^2-4y+7\)

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

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

Mà \(\left(x+y-5\right)^2\ge0\forall x;y\in R\)

\(\left(y-1\right)^2\ge0\forall y\in R\Rightarrow2\left(y-1\right)^2\ge0\forall y\in R\)

\(\Rightarrow-A\ge5\)

\(\Leftrightarrow A\le-5\)

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 Max A = - 5 khi ( x ; y ) = ( 4 ; 1 )