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D=3y^2 -2y+10
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![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề:
\(C=x^2-4xy+5y^2-10y+6\)
\(C=\left(x^2-4xy+4y^2\right)+\left(y^2-10y+25\right)-19\)
\(C=\left(x-2y\right)^2+\left(y-5\right)^2-19\ge-19\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-2y\right)^2=0\\\left(y-5\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2y\\y=5\end{cases}}\Rightarrow\hept{\begin{cases}x=10\\y=5\end{cases}}\)
Vậy \(Min_C=-19\Leftrightarrow\hept{\begin{cases}x=10\\y=5\end{cases}}\)
\(D=x^2-2xy+2y^2-2x-10y+20\)
\(D=\left(x-y\right)^2-2\left(x-y\right)+1+\left(y^2-12y+36\right)-17\)
\(D=\left(x-y-1\right)^2+\left(y-6\right)^2-17\ge-17\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-6\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=y+1\\y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\y=6\end{cases}}\)
Vậy \(Min_D=-17\Leftrightarrow\hept{\begin{cases}x=7\\y=6\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2+y^2-2xy-2x+2y=\left(x^2+y^2+1-2xy-2x+2y\right)+3x^2.\)
\(=\left(x-y-1\right)^2+3x^2\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\3x^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải
\(C=\left(x+1\right)^2+\left(1-2y\right)^2+5\)
Vì \(\left(x+1\right)^2\ge0\) Dấu " = " xảy ra khi \(\left(x+1\right)^2=0\text{ }\Rightarrow\text{ }x+1=0\text{ }\Rightarrow\text{ }x=-1\)
\(\left(1-2y\right)^2\ge0\)Dấu " = " xảy ra khi \(\left(1-2y\right)^2=0\text{ }\Rightarrow\text{ }1-2y=0\text{ }\Rightarrow\text{ }2y=1\text{ }\Rightarrow\text{ }y=\frac{1}{2}\)
\(\Rightarrow\text{ }C=\left(x+1\right)^2+\left(1-2y\right)^2+5\ge0+0+5\ge5\)
\(\Rightarrow\text{ Min C = }5\text{ khi }x=-1\text{ , }y=\frac{1}{2}\)
Bài giải
\(D=-277-\left(x-y\right)^2-\left|3y+9\right|\)
Vì \(\left(x-y\right)^2\ge0\) Dấu " = " xảy ra khi \(\left(x-y\right)^2=0\text{ }\Rightarrow\text{ }x-y=0\text{ }\Rightarrow\text{ }x=y\)
\(\left|3y+9\right|\ge0\text{ }\) Dấu " = " xảy ra khi \(\left|3y+9\right|=0\text{ }\Rightarrow\text{ }3y+9=0\text{ }\Rightarrow\text{ }3y=-9\text{ }\Rightarrow\text{ }y=-9\text{ : }3\text{ }\Rightarrow\text{ }y=-3\)
\(\Rightarrow\text{ }x=y=-3\)
\(\Rightarrow\text{ }B=-277-\left(x-y\right)^2-\left|3y+9\right|\le-277-0-0=-277\)
\(\Rightarrow\text{ }\text{Max D = }-277\text{ khi }x=y=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=x^3\left(x-y\right)-y^3\left(x-y\right)+\left(x^2y^2-8xy+16\right)+1984\)
\(D=\left(x-y\right)\left(x^3-y^3\right)+\left(xy-4\right)^2+1984\)
\(D=\left(x-y\right)^2\left(x^2+xy+y^2\right)+\left(xy-4\right)^2+1984\)
\(D=\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]+\left(xy-4\right)^2+1984\ge1984\)
\(D_{min}=1984\) khi \(x=y=\pm2\)
\(D=3y^2-2y+10=3\left(y^2-\frac{2}{3}y+\frac{10}{3}\right)\)
\(=3\left(y^2-\frac{2}{3}y+\frac{1}{9}+\frac{29}{9}\right)\)
\(=3\left(y-\frac{1}{3}\right)^2+\frac{29}{3}\ge\frac{29}{3}\)
=> Min D = 29/3
Dấu "=" xảy ra <=> \(y-\frac{1}{3}=0\Rightarrow y=\frac{1}{3}\)
Vậy Min D = 29/3 khi y = 1/3
\(\exists maxD\)