\(N=2x^2+y^2+2xy-2x-2y+2011\)
\(=\left(x^2+y^2+2xy\right)-2\left(x+y\right)+1+x^2+2010\)
\(=\left(x+y-1\right)^2+x^2+2010\ge2010\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\)
Vậy Min N là : \(2010\Leftrightarrow x=0;y=1\)
\(P=2x\left(1-x\right)=2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\forall x\)Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy Max P là : \(\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)
\(Q=-x^2-4y^2+4x+2y-25\)
\(=-\left(x^2-4x+4\right)-\left(4y^2-2y+\dfrac{1}{4}\right)-\dfrac{83}{4}\)
\(=-\left(x-2\right)^2-\left(2y-\dfrac{1}{2}\right)^2-\dfrac{83}{4}\le\dfrac{83}{4}\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2y-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{4}\end{matrix}\right.\)
Vậy Max Q là : \(\dfrac{83}{4}\Leftrightarrow x=2;y=\dfrac{1}{4}\)
P=2x(1-x) đấy nhé,chứ ko phải như trên đâu
Tìm giá trị nhỏ nhất của biểu thức N
Giải:
Ta có:
\(N=2x^2+y^2+2xy-2x-2y+2011\)
\(\Rightarrow N=y^2+\left(2xy-2y\right)+2x^2-2x+2011\)
\(\Rightarrow N=y^2+2y\left(x+1\right)+\left(x+1\right)^2+2x^2-2x+2011-\left(x+1\right)^2\)
\(\Rightarrow N=\left(y+x+1\right)^2+2x^2-2x-\left(x^2+2x+1\right)\)
\(\Rightarrow N=\left(y+x+1\right)^2+x^2-4x-1\)
\(\Rightarrow N=\left(y+x+1\right)^2+x^2-4x+4-5\)
\(\Rightarrow N=\left(y+x+1\right)^2+\left(x-2\right)^2\)
Vì \(\left(y+x+1\right)^2\ge0;\left(x-2\right)^2\ge0\) nên \(N\ge-5\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y+x+1=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy minN = -5 <=> x = 2; y = -3
Bạn kiểm tra là biểu thức Q nhé. Nó bị sai rồi
