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ta có \(2B=2x^2-4xy+4y^2+10x\)
\(=\left(x^2-4xy+4y^2\right)+\left(x^2+10x+25\right)-25\)
\(=\left(x-2y\right)^2+\left(x+5\right)^2-25\)
vì \(\left(x-2y\right)^2>=0;\left(x+5\right)^2>=0\)
=>\(2B>=-25=>b>=-\frac{25}{2}\)
dấu = xảy ra <=> \(\hept{\begin{cases}x=-5\\y=-10\end{cases}}\)
b) ta có
\(Q=x^2-6xy+9y^2+x^2-x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-3y\right)^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
=> Q>=3/4
dấu = xảy ra <=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{2}\end{cases}}\)
\(C=\left(x^2-6xy+9y^2\right)+\left(x^2-2x+1\right)+2017=\left(x-3y\right)^2+\left(x-1\right)^2+2017\)
\(\ge0+0+2017=2017.\Rightarrow C_{min}=2017\Leftrightarrow\hept{\begin{cases}x-1=0\\x-3y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\end{cases}}\)
\(A=x^2-2xy+6y^2-12x+2y+54\)
\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)
\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)
Do: \(\left(x-y-6\right)^2\ge0\forall xy\); \(5\left(y-1\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-y-6\right)^2+5\left(y-1\right)^2\ge0\)
\(\Leftrightarrow A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
\(\Rightarrow A_{Min}=4\)
Dấu "=" xảy ra khi \(x=7;y=1\)
\(A=x^2-2xy+6y^2-12x+2y+45\)
\(=x^2+y^2+36-2xy-12x+12y+5y^2-10y+5+4\)
\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
Gía trị nhỏ nhất : \(A=4\)Khi \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=7\end{cases}}\)
\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right)\\ \)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)\(\ge4\)
Amin=4 khi y=1; x=7
\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right) \)
\(A=\left(x-7-6\right)^2+5\left(y-1^2\right)+4\ge4\)
\(Amin=4\)\(khi\)\(y=1;x=7\)
\(C=2x^2+9y^2-6xy-2x+2018\)
\(=\left(x^2-6xy+9y^2\right)+\left(x^2-2x+1\right)+2017\)
\(=\left(x-3y\right)^2+\left(x-1\right)^2+2017\)
Nhận xét :
\(\left\{{}\begin{matrix}\left(x-3y\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2+2017\ge2017\)
\(\Leftrightarrow C\ge2017\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(C_{Min}=2017\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(D=x^2-2xy+6y^2-12x+2y+45\)
\(=\left(x^2-2xy+y^2\right)-\left(12x+12y\right)-10y+5y^2+45\)
\(=\left(x-y\right)^2-12\left(x-y\right)+36+\left(5y^2-10y+5\right)+4\)
\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)
Nhận xét :
\(\left\{{}\begin{matrix}\left(x-y-6\right)^2\ge0\\5\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x-y-6\right)+5\left(y-1\right)^2+4\ge4\)
\(\Leftrightarrow D\ge4\)
Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
Vậy \(D_{Min}=4\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)