a) \(A=x^2-4x+6\)
\(A=x^2-2\cdot x\cdot2+4+2\)
\(A=\left(x-2\right)^2+2\)
Mà: \(\left(x-2\right)^2\ge0\) nên \(A=\left(x-2\right)^2+2\ge2\)
Dấu "=" xảy ra:
\(\left(x-2\right)^2+2=2\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy: \(A_{min}=2\) khi \(x=2\)
a)
\(A=x^2-4x+6\\ =x^2-4x+4+2\\ =\left(x-2\right)^2+2\ge2\)
GTNN của A đạt 2 khi x = 2
b)
\(B=y^2-y+1\\ =y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
GTNN của B đạt \(\dfrac{3}{4}\) khi \(y=\dfrac{1}{2}\)
c)
\(C=x^2-4x+y^2-y+5\\ =\left(x^2-2.2x+4\right)+\left(y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
GTNN của C đạt \(\dfrac{3}{4}\) khi x = 2 và y = \(\dfrac{1}{2}\)
b) \(B=y^2-y+1\)
\(B=y^2-2\cdot\dfrac{1}{2}\cdot y+\dfrac{1}{4}+\dfrac{3}{4}\)
\(B=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà: \(\left(y-\dfrac{1}{2}\right)^2\ge0\) nên \(B=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra:
\(\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow y-\dfrac{1}{2}=0\)
\(\Leftrightarrow y=\dfrac{1}{2}\)
Vậy: \(B_{min}=\dfrac{3}{4}\) khi \(y=\dfrac{1}{2}\)
a, A= (x^2-2.2x+4)+2=(x-2)^2+2
ta có (x-2)^2 luôn lớn hơn bằng 0 =>Aluoon lớn hơn bằng 0+2
dấu = xảy ra khi x=2
b,
3434 khi C=x2−4x+y2−y+5=(x2−2.2x+4)+(y2−2.12y+14)+34=(x−2)2+(y−12)2+34≥34�=�2−4�+�2−�+5=(�2−2.2�+4)+(�2−2.12�+14)+34=(�−2)2+(�−12)2+34≥34
GTNN của C đạt 12
c) \(C=x^2-4x+y^2-y+5\)
\(C=x^2-4x+4+y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\)
\(C=\left(x^2-4x+4\right)+\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(C=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-\dfrac{1}{2}\right)^2\ge0\end{matrix}\right.\) nên:
\(C=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(C_{min}=\dfrac{3}{4}\) khi \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
