2P = \(2x^2+4xy+4y^2-12x-8y+50\)

      = \(\left(x+2y\right)^2-2\left(x+2y\right)\cdot2+4+x^2-8x+16+30\)

      = \(\left(x+2y-2\right)^2+\left(x-4\right)^2+30\ge30\)

=> P \(\ge15\)

Dấu '' = '' xảy ra khi x = 4 ; y = -1

P = x² + 2y² + 2xy - 6x - 4y + 25 đạt GTNN khi x² + 2y² + 2xy - 6x - 4y = -25 và P = 0

Lập luận đỉnh cao!! ^~^

\(Q=\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy+2016=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{4xy}+4xy+\frac{5}{4xy}+2016\)

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\). Dấu "=" khi a=b (bạn tự chứng minh)

\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}=4\)

Vì x>0, y>0 nên xy>0

Áp dụng bất đẳng thức Cô si cho 2 số dương

\(\frac{1}{4xy}+4xy\ge2\sqrt{\frac{1}{4xy}.4xy}=2\)

Ta có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow\frac{5}{4xy}\ge5\)

Dấu "=" khi \(\hept{\begin{cases}x^2+y^2=2xy\\\frac{1}{4xy}=4xy\\x=y\end{cases}\Rightarrow x=y=\frac{1}{2}}\)

\(\Rightarrow Q\ge4+2+5+2016=2027\)

Vậy \(minQ=2027\)khi \(x=y=\frac{1}{2}\)

\(M=\frac{2x^2+4xy+2y^2+8xy}{x+y}=\frac{2\left(x^2+2xy+y^2\right)+2\cdot4xy}{x+y}=\frac{2\left(x+y\right)^2+2\cdot1}{x+y}\)

\(=2\left(x+y\right)+\frac{2}{x+y}>=2\sqrt{2\left(x+y\right)\cdot\frac{2}{x+y}}=2\cdot\sqrt{4}=2\cdot2=4\)(bđt cosi)

dấu = xảy ra khi x=y=\(\frac{1}{2}\)

vậy min M là 4 khi \(x=y=\frac{1}{2}\)

2) ĐKXĐ:  \(1\le x\le5\)

\(B^2=\left(\sqrt{x-1}+\sqrt{5-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+5-x\right)=8\Rightarrow B\le2\sqrt{2}\)

Xảy ra đẳng thức khi và chỉ khi x = 3

\(x^2+5y^2+9z^2-4xy-6yz+12\)

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

\(=\left(x-2y\right)^2+\left(y-3z\right)^2+12\ge12\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2y=0\\y-3z=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2y\\y=3z\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=6z\\y=3z\end{cases}}\)

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

\(A=4-\left(2x-y\right)^2-\left(y-1\right)^2\le4\)

\(A_{max}=4\) khi \(\hept{\begin{cases}x=\frac{1}{2}\\y=1\end{cases}}\)

Chúc bạn học tốt !!!

\(-4x^2+4xy-2y^2+2y+3\)

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

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

Ta có \(\left(2x+y\right)^2\ge0\)  \(\forall x,y\) \(;\left(y-1\right)^2\ge0\)  \(\forall y\)

=> \(\left(2x+y\right)^2+\left(y-1\right)^2\ge0\)   \(\forall x,y\)

=> \(-\left(2x+y\right)^2-\left(y-1\right)^2\le0\)  \(\forall x,y\)

=> \(-\left(2x+y\right)-\left(y-1\right)^2+4\le4\)  \(\forall x,y\)

\(MaxA=4\Leftrightarrow\hept{\begin{cases}\left(y-1\right)^2=0\\\left(2x+y\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}y-1=0\\2x+y=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\x=-\frac{1}{2}\end{cases}}}\)

\(a,A=3x^2-5x+1\)

\(=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}\)

\(=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\)

Với mọi giá trị của x ta có:

\(\left(x-\dfrac{5}{6}\right)^2\ge0\)

\(\Rightarrow3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)

Vậy Min \(A=-\dfrac{13}{12}\)

Để \(A=-\dfrac{13}{12}\) thì \(x-\dfrac{5}{6}=0\Rightarrow x=\dfrac{5}{6}\)

\(b,B=2x^2+5y^2-4x+2y+4xy+2017\)

\(=\left(2x^2-4x+4xy\right)+5y^2+2y+2017\)

\(=2\left(x^2-2x+2xy\right)+5y^2+2y+2017\)

\(=2\left[x^2-2x\left(1-y\right)+\left(1-y\right)^2\right]+5y^2+2y+2017+2\left(1-y\right)^2\)\(=2\left(x-1+y\right)^2+5y^2+2y+2017-2\left(1-y\right)^2\)

\(=2\left(x+y-1\right)^2+5y^2+2y+2017-2+4y-2y^2\)\(=2\left(x+y-1\right)^2+3y^2+6y+2015\)

\(=2\left(x+y-1\right)^2+3\left(y^2+2y+1\right)+2012\)

\(=2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\)

Với mọi giá trị của x ta có:

\(2\left(x+y-1\right)^2\ge0;3\left(y+1\right)^2\ge0\)

\(\Rightarrow2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\ge2012\) Vậy : Min B = 2012

Để B = 2012 thì \(\left\{{}\begin{matrix}x+y-1=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)