a: A=x^2-2xy+y^2+y^2-4y+4+1

=(x-y)^2+(y-2)^2+1>=1
Dấu = xảy ra khi x=y=2

b: B=4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1-2

=(2x+2y)^2+(x-1)^2+(y+1)^2-2>=-2

Dấu = xảy ra khi x=1 và y=-1

\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

                                             \(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)

Dấu "=" <=> x= y = 1/2

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

bài 3 min hay max ?

Thiếu đề !

Chuyển biểu thức , ta có :

x² - 8xy + 20y² - 4y + 1

= x² - 8xy + 16y² + 4y² - 4y + 1

= (x² - 8xy + 16y²) + (4y² - 4y + 1)

= (x - 4y)² + (2y + 1)² 

Còn lại do thiếu đề nên không thể làm tiếp 

Ah, mình đánh thiếu mất  đề là \(x^2-8xy+20y^2-4y+1=0\)

\(x^2+y^2-2x-4y-4=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2-9=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=9=0^2+3^2=0^2+\left(-3\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y-2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=3\\y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=0\\y-2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-3\\y-2=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow-2\le x\le4\left(y\in R\right)\)

Ta có \(S=3x+4y\)

Mà \(x\ge-2;y\ge-1\Leftrightarrow S\ge3\cdot\left(-2\right)+4\cdot\left(-1\right)=-6-4=-10\)

Vậy GTNN của S là \(-10\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

Lời giải:

ĐKĐB $\Leftrightarrow (x^2-2x+1)+(y^2-4y+4)-9=0$

$\Leftrightarrow (x-1)^2+(y-2)^2-9=0$

$\Rightarrow (x-1)^2=9-(y-2)^2\leq 9$

$\Rightarrow -3\leq x-1\leq 3$

$\Leftrightarrow -2\leq x\leq 4$

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Đặt $x-1=a; y-2=b$ thì bài toán trở thành:
Cho $a,b$ thực thỏa mãn $a^2+b^2=9$

Tìm min $S=3a+4b+11$

Áp dụng BĐT Bunhiacopxky:

$(3a+4b)^2\leq (a^2+b^2)(3^2+4^2)=9.25$

$\Rightarrow -15\leq 3a+4b\leq 15$

$\Rightarrow 3a+4b\geq -15$

$\Rightarrow S=3a+4b+11\geq -4$

Vậy $S_{\min}=-4$ khi $x=\frac{-4}{5}; y=\frac{-1}{5}$

câu A thiếu đề

B=\(x^2-2x+2017=\left(x-1\right)^2+2016>=2016\)

Min B=2016 khi x-1=0<=>x=1

+)D=\(-2x^2+4x+2017=-2\left(x^2-2x+1\right)+2019=-2\left(x-1\right)^2+2019< =2019\)

=>Max D=2019, dấu '=' xảy ra khi x-1=0<=>x=1

Bổ sung câu A. \(A=x^2+2xy+3y^2-4y+2017\)

\(A=x^2+2xy+3y^2-4y+2017\)

\(A=\left(x^2+2xy+y^2\right)+\left(2y^2-4y+2\right)+2015\)

\(A=\left(x^2+2xy+y^2\right)+2\left(y^2-2y+1\right)+2015\)

\(A=\left(x+y\right)^2+2\left(y-1\right)^2+2015\ge2015\)

Vậy A_min=2015 <=> x=-1 và y=1

\(x^2+y^2+\frac{8xy}{x+y}=16\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y\right)+8xy-16\left(x+y\right)=0\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y-4\right)+4x^2+4y^2+8xy-16\left(x+y\right)=0\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y-4\right)+4\left(x+y\right)^2-16\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y-4\right)\left(x^2+y^2+4x+4y\right)=0\)

\(\Leftrightarrow x+y-4=0\)(vì \(x^2+y^2+4x+4y>0\))

\(\Leftrightarrow y=4-x\).

\(Q=x^2-2x+4y+100=x^2-2x+4\left(4-x\right)+100\)

\(=x^2-6x+116=\left(x-3\right)^2+107\ge107\)

Dấu \(=\)khi \(x=3\Rightarrow y=1\).

\(2P=2x^2+8y^2+\dfrac{150}{x}+\dfrac{2}{y}\)

\(=\dfrac{7}{5}x^2+7y^2+\left(\dfrac{3}{5}x^2+\dfrac{75}{x}+\dfrac{75}{x}\right)+\left(y^2+\dfrac{1}{y}+\dfrac{1}{y}\right)\) 

Ta có: \(\left(5+1\right)\left(x^2+5y^2\right)\ge5\left(x+y\right)^2\Rightarrow\dfrac{7\left(x^2+5y^2\right)}{5}\ge\dfrac{7\left(x+y\right)^2}{6}\ge42\)

\(\Rightarrow2P\ge42+3\sqrt[3]{\dfrac{3.75^2.x^2}{5x^2}}+3\sqrt[3]{\dfrac{y^2}{y^2}}=90\)

\(\Rightarrow P\ge45\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(5;1\right)\)