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Ta có :
\(B=x^2+xy+y^2-2x-3y+2019\)
\(\Leftrightarrow4B=4x^2+4xy+4y^2-8x-12y+8076\)
\(\Leftrightarrow4B=\left(4x^2+4xy+y^2\right)-4\left(2x+y\right)+4+3y^2-4y+4022\)
\(\Leftrightarrow2B=\left(2x+y\right)^2-4\left(2x+y\right)+4+3\left(y^2-\frac{4}{3}y+\frac{4}{9}\right)+\frac{12062}{3}\)
\(\Leftrightarrow2B=\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2+\frac{12062}{3}\ge\frac{12062}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{2}{3}\)
Bạn kiểm tra lại nhé, mình k chắc có đúng k nữa !
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\(A=x^2+2y^2-2xy-2y-2x+2019\)
\(A=x^2+y^2+y^2-2xy+2y-4y-2x+2019\)
\(A=\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+y^2-4y+4+2014\)
\(A=\left(x-y\right)^2-2\left(x-y\right)+1+\left(y-2\right)^2+2014\)
\(A=\left(x-y-1\right)^2+\left(y-2\right)^2+2014\ge2014\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2-1=0\\y=2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}}\)
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\(4B=4x^2+4xy+4y^2-8x-12y+8076\)
= \(\left(2y\right)^2-4y\left(3-x\right)+\left(3-x\right)^2-\left(3-x\right)^2\)
\(+\left(2x\right)^2-8x+8076\)
= \(\left(2y-3+x\right)^2+3x^2-2x+8076\)
đến đây thì dễ rồi
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\(A=x^2+y^2+1-2xy-2x+2y+y^2-4y+4+2014\)
\(=\left(x-y-1\right)^2+\left(y-2\right)^2+2014\ge2014\)
\(\Rightarrow A_{min}=2014\) khi \(\left\{{}\begin{matrix}y-2=0\\x-y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)
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a) ... = (x^2 -2xy + y^2)+(x^2 -2x+1)+2014=(x-y)^2 + (x-1)^2 +2014 >= 2014
Đăngt thức xay ra khi x=y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(H=x^2+2xy+y^2+2x+2y+x^2+4x+2019=\left(x+y\right)^2+2\left(x+y\right)+\left(x+2\right)^2+2015\)
\(=\left(x+y+1\right)^2+\left(x+2\right)^2+2014\ge2014\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-2;y=1\)
\(I=\left(1-x\right)^2+\left(-2-y\right)^2+\left(x+y\right)^2\ge\frac{\left(1-x-2-y+x+y\right)^2}{3}=\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(1-x=-2-y=x+y\)\(\Leftrightarrow\)\(x=\frac{4}{3};y=\frac{-5}{3}\)
\(B=x^2+\frac{y^2}{4}+1+xy-2x-y+\frac{3}{4}\left(y^2-\frac{4}{3}y+\frac{4}{9}\right)+\frac{6056}{3}\)
\(B=\left(x+\frac{y}{2}-1\right)^2+\frac{3}{4}\left(y-\frac{2}{3}\right)^2+\frac{6056}{3}\ge\frac{6056}{3}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{2}{3}\\y=\frac{2}{3}\end{matrix}\right.\)