Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(A_{MIN}\Leftrightarrow\hept{\begin{cases}x+y+2017=2013\\y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=0\end{cases}}\)
\(A_{MAX}\Leftrightarrow\hept{\begin{cases}x+y+2017=2015\\y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=0\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)
Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)
\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)
\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2xy+6x+6y+2y^2+8=0\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)
\(\Leftrightarrow\left(x+y+3\right)^2=1-y^2\)
Ta thấy : \(1-y^2\le1\forall y\) \(\Rightarrow\left(x+y+3\right)^2\le1\)
\(\Rightarrow-1\le x+y+3\le1\)
\(\Rightarrow-1+2013\le x+y+3+2013\le1+2013\)
\(\Rightarrow2012\le x+y+2016\le2014\)
Vậy ta có :
+) Min \(B=2012\) . Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-4\end{cases}}\)
+) Max \(M=2014\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-2\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2xy+6x+6y+2y^2+8=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(6x+6y\right)+9+y^2-1=0\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)
\(\left(x+y+3\right)^2=1-y^2\)
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 \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+y+3\right)^2=1-y^2\)
Ta thấy \(1-y^2\le1\) do \(y^2\ge0\forall y\)
Suy ra \( \left(x+y+3\right)^2\le1\Rightarrow\left|x+y+3\right|\le1\Rightarrow-1\le x+y+3\le1\)
\(\Rightarrow2012\le x+y+2016\le2014\)
\(Min_{\left(B\right)}=2012\Leftrightarrow x=-4;y=0\)
\(Max_{\left(B\right)}=2014\Leftrightarrow x=-2;y=0\)
Chúc bạn học tốt !!!
Lời giải:
$x^2-2xy+6y^2-12x+2y+41=0$
$\Leftrightarrow (x^2-2xy+y^2)+5y^2-12x+2y+41=0$
$\Leftrightarrow (x-y)^2-12(x-y)+36+5y^2-10y+5=0$
$\Leftrightarrow (x-y-6)^2+5(y-1)^2=0$
Vì $(x-y-6)^2\geq 0; (y-1)^2\geq 0$ với mọi $x,y$
Do đó để tổng trên bằng $0$ thì bản thân mỗi số trên bằng $0$
$\Rightarrow x-y-6=y-1=0$
$\Rightarrow y=1; x=7$
$\Rightarrow P=2021(10-7-2)^{2021}-8(6-7)^{2022}$
$=2021-8=2013$