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\(x^2+y^2=6x-5\)
\(\left(x-3\right)^2+y^2=2^2\Rightarrow1\le x\le5\)
\(1\le x^2+y^2\le25\)
Có xy ≤ 1/4 (x+y)^2
=> 3xy ≤ 3/4 (x+y)^2
=> T = x^2-xy+y^2 = (x+y)^2 - 3xy ≥ (x+y)^2 - 3/4 (x+y)^2 = 1/4 (x+y)^2
=10201/4
Dấu = xảy ra khi x=y=101/2
T = (x+y)^2 - 3xy <= (x+y)^2 = 101^2 = 10201
Dấu = xảy ra khi 1 số = 0, 1 số = 101
Giải:
Đặt \(A=x+y+2017\) Ta có: \(x^2+2xy+6x+6y+2y^2+8=0\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)
Mà \(y^2\ge0\Rightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\) \(\Leftrightarrow\left(x+y+3\right)^2\le1\)
\(\Rightarrow\left|x+y+3\right|\le1\Rightarrow-1\le x+y+3\le1\)
\(\Leftrightarrow2013\le A\le2015\) Dấu "=" xảy ra:
\(A_{MIN}\Leftrightarrow\hept{\begin{cases}x+y+2017=2013\\y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=0\end{cases}}\)
\(A_{MAX}\Leftrightarrow\hept{\begin{cases}x+y+2017=2015\\y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=0\end{cases}}\)
\(x^2+2xy+6x+6y+2y^2+8=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(6x+6y\right)+9+y^2-1=0\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)
\(\left(x+y+3\right)^2=1-y^2\)
Do \(VP=1-y^2\le1\forall x\) \(\Rightarrow VT=\left(x+y+3\right)^2\le1\)
\(\Leftrightarrow-1\le x+y+3\le1\)
\(\Leftrightarrow-1+2013\le x+y+3+2013\le1+2013\)
\(\Leftrightarrow2012\le x+y+2016\le2014\) hay \(2012\le B\le2014\)
B đạt MIN là 2012 \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Rightarrow\hept{\begin{cases}y=0\\x=-4\end{cases}}}\)
B đạt MAX là 2014 \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}}\)
\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)
Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)
\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)
\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)
Ta có : (x+y)2+7x+7y+y2+6=0
( x2 + y2 + \(\frac{49}{4}\)+ 7x + 7y + 2xy ) + y2 - \(\frac{25}{4}\)= 0
( x + y + \(\frac{7}{2}\))2 = \(\frac{25}{4}\)- y2 \(\le\frac{25}{4}\)
\(\Rightarrow\frac{-5}{4}\le x+y+\frac{7}{2}\le\frac{5}{4}\)
\(\Rightarrow\frac{-15}{4}\le x+y+1\le\frac{-5}{4}\)
\(\Rightarrow\)......
lon so roi,
thay -5/4 thành -5/2 ; 5/4 thành 5/2
-15/4 thành -5 ; 5/2 thành 0
