Mình chỉ tìm giá trị chứ không tìm x đâu nhé (đề bài ghi thế)
a)
\(A=x^2-6x+11\\ =x^2-6x+9+2\\ =\left(x-3\right)^2+2\)
\(\left(x-3\right)^2\ge0\forall x\\ 2\ge2\\ \Rightarrow\left(x-3\right)^2+2\ge2\forall x\\ A\ge2\forall x\\ \Rightarrow A_{min}=2\)
b) B = 2x2 + 10 - 1
B = 2(x2 + 5) - 1
B = 2(x2 + 2.\(\frac{5}{2}\).x + \(\frac{25}{4}\)) - \(\frac{25}{2}\) - 1
B = 2(x + \(\frac{5}{2}\))2 - \(\frac{27}{2}\)
Vậy GTNN của B = \(\frac{-27}{2}\) khi x = \(\frac{-5}{2}\).
c) C = 5x - x2
C = -(x2 - 5x)
C = -(x2 - 2.\(\frac{5}{2}\).x + \(\frac{25}{4}\)) + \(\frac{25}{4}\)
C = -(x - \(\frac{5}{2}\))2 + \(\frac{25}{4}\)
Vậy GTLN của C = \(\frac{25}{4}\) khi x = \(\frac{5}{2}\).
a) \(A=x^2-6x+11\)
\(\Leftrightarrow A=x^2-6x+9+2\)
\(\Leftrightarrow A=\left(x^2-2.x.3+3^2\right)+2\)
\(\Leftrightarrow A=\left(x-3\right)^2+2\)
Vậy GTNN của \(A=2\) khi \(x-3=0\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(\Leftrightarrow B=2x^2+10x+\dfrac{25}{2}-\dfrac{27}{2}\)
\(\Leftrightarrow B=\left(2x^2+10x+\dfrac{25}{2}\right)-\dfrac{27}{2}\)
\(\Leftrightarrow B=2\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{27}{2}\)
\(\Leftrightarrow B=2\left[x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{27}{2}\)
\(\Leftrightarrow B=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\)
Vậy GTNN của \(B=\dfrac{-27}{2}\) khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{-5}{2}\)
c) \(C=5x-x^2\)
\(\Leftrightarrow C=-x^2+5x-\dfrac{25}{4}+\dfrac{25}{4}\)
\(\Leftrightarrow C=-\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(\Leftrightarrow C=-\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{25}{4}\)
\(\Leftrightarrow C=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)
Vậy GTLN của \(C=\dfrac{25}{4}\) khi \(x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)
