1: \(M=x^2-3x+5\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
2 Ta có: \(A=x^2-x+3\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{11}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
1) \(M=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
\(minM=\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{2}\)
2) \(A=x^2-x+3=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{11}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
\(minA=\dfrac{11}{4}\Leftrightarrow x=\dfrac{1}{2}\)
3) \(N=x^2+y^2-3x+2y+3=\left(x^2-3x+\dfrac{9}{4}\right)+\left(y^2+2y+1\right)+\dfrac{1}{4}=\left(x-\dfrac{3}{2}\right)^2+\left(y+1\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
\(minN=\dfrac{1}{4}\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-1\end{matrix}\right.\)
4) \(B=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+4y+4\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+2\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(minB=\dfrac{3}{4}\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-2\end{matrix}\right.\)
5) \(P=2\left[x^2+2x\left(y+2\right)+\left(y+2\right)^2\right]+3\left(y^2-4y+4\right)-5=2\left(x+y+2\right)^2+3\left(y-2\right)^2-5\ge-5\)
\(minP=-5\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
6) \(C=2x^2+4y^2+4xy-3x-1=\left(x^2+4xy+4y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{13}{4}=\left(x+2y\right)^2+\left(x-\dfrac{3}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\)
\(minC=-\dfrac{13}{4}\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-\dfrac{3}{4}\end{matrix}\right.\)
