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D ez nhất :v
\(D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+5\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+5\ge5\)
Đẳng thức xảy ra khi x = 1 và y = -2
\(A=\left[\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4\right]+\left(y^2-2y+1\right)+2020\)
\(=\left[\left(x-y\right)^2+2\left(x-y\right).2+2^2\right]+\left(y-1\right)^2+2020\)
\(=\left(x-y+2\right)^2+\left(y-1\right)^2+2020\ge2020\)
Dấu "=" xảy ra khi y = 1 và x - y + 2 = 0 tức là x = y - 2 = -1
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
\(G=x^2-2xy+2y^2+2x-10y+17\\ \\ =x^2-2xy+y^2+y^2+2x-2y-8y+1+16\\ \\ =\left(x^2+y^2+1-2xy+2x-2y\right)+\left(y^2-8y+16\right)\\ \\ =\left(x-y+1\right)^2+\left(y-4\right)^2\)
Do \(\left(x-y+1\right)^2\ge0\forall x;y\)
\(\left(y-4\right)^2\ge0\forall y\)
\(\Rightarrow G=\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\forall x;y\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy \(G_{\left(Min\right)}=0\) khi \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
\(H=x^2+2xy+y^2-2x-2y\\ =x^2+2xy+y^2-2x-2y+1-1\\ =\left(x^2+y^2+1+2xy-2x-2y\right)-1\\ \\ =\left(x+y-1\right)^2-1\)
Do \(\left(x+y-1\right)^2\ge0\forall x;y\)
\(\Rightarrow H=\left(x+y-1\right)^2-1\ge-1\forall x;y\)
Dấu \("="\) xảy ra khi:
\(\left(x+y-1\right)^2=0\\ \Leftrightarrow x+y-1=0\\ \Leftrightarrow x+y=1\)
Vậy \(H_{\left(Min\right)}=-1\) khi \(x+y=1\)
\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)