Answer:

3.

\(x^2+2y^2+2xy+7x+7y+10=0\)

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

\(\Rightarrow\left(x+y\right)^2+7.\left(x+y\right)+y^2+10=0\)

\(\Rightarrow4S^2+28S+4y^2+40=0\)

\(\Rightarrow4S^2+28S+49+4y^2-9=0\)

\(\Rightarrow\left(2S+7\right)^2=9-4y^2\le9\left(1\right)\)

\(\Rightarrow-3\le2S+7\le3\)

\(\Rightarrow-10\le2S\le-4\)

\(\Rightarrow-5\le S\le-2\left(2\right)\)

Dấu " = " xảy ra khi: \(\left(1\right)\Rightarrow y=0\)

Vậy giá trị nhỏ nhất của \(S=x+y=-5\Rightarrow\hept{\begin{cases}y=0\\x=-5\end{cases}}\)

Vậy giá trị lớn nhất của \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

Đặt \(A=-2x^2-y^2-2xy+4x+2y+2\)

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

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

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)-4\)

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

Mà  \(\left(x+y-1\right)^2\ge0\forall x;y\)

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

\(\Rightarrow-A\ge-4\)

\(\Leftrightarrow A\le4\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)

Vậy  \(A_{Max}=4\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Đặt  \(B=x^2-4xy+5y^2+10x-22y+27\)

\(B=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+27\)

\(B=\left[\left(x-2y\right)^2+2\left(x-2y\right)\times5+25\right]+\)\(\left(y^2-2y+1\right)+1\)

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

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

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

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy  \(B_{Min}=1\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

ta có:

\(P=\frac{2x^2-2xy+9y^2}{x^2+2xy+5y^2}=1+\frac{\left(x-2y\right)^2}{x^2+2xy+5y^2}=\frac{17}{4}-\frac{1}{3}.\frac{\left(3x+7y\right)^2}{x^2+2xy+5y^2}\)

\(\Rightarrow\hept{\begin{cases}min_P=1\\max_P=\frac{17}{4}\end{cases}}\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\Rightarrow\dfrac{y}{x}\ge4\)

\(P=\dfrac{1-\dfrac{2y}{x}+2\left(\dfrac{y}{x}\right)^2}{1+\dfrac{y}{x}}\)

Đặt \(\dfrac{y}{x}=a\ge4\Rightarrow P=\dfrac{2a^2-2a+1}{a+1}=2a-4+\dfrac{5}{a+1}\)

\(P=\dfrac{a+1}{5}+\dfrac{5}{a+1}+\dfrac{9}{5}.a-\dfrac{21}{5}\ge2\sqrt{\dfrac{5\left(a+1\right)}{5\left(a+1\right)}}+\dfrac{9}{5}.4-\dfrac{21}{5}=5\)

Dấu "=" xảy ra khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

Giải :(x²+2xy+y²)+y²-6x-8y+2024=(x+y)²-2(x+y)3+y²-2y+2024

=(x+y-3)²+(y²-2y+1)+2014=(x+y-3)²+(y-1)²+2014 >=2014

vì (x+y-3)²;(y-1)²>=0 với mọi x;y

nên Pmin=2014khi y=1;x=2

MinP=2024 nha!

2024 đó !đúng 100% luôn !