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a) \(A=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\)
\(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+2\ge2\)
Đẳng thức xảy ra <=> x - 3 = 0 => x = 3
Vậy AMin = 2 , đạt được khi x = 3
b) \(B=5x-x^2=-x^2+5x=-x^2+5x-\frac{25}{4}+\frac{25}{4}=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
\(-\left(x-\frac{5}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2
Vậy BMax = 25/4 , đạt được khi x = 5/2
c) \(2x-2x^2-5=-2x^2+2x-5=-2\left(x^2-x+\frac{1}{4}\right)-\frac{9}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)
\(-2\left(x-\frac{1}{2}\right)^2\le0\forall x\Rightarrow-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le-\frac{9}{2}\)
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
Vậy CMax = -9/2 , đạt được khi x = 1/2
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
\(a,A=x^2-6x+11=\left(x-3\right)^2+2\)\(\Leftrightarrow Amin=2\)
Dấu = xảy ra \(\Leftrightarrow x=3\)
\(2x^2+10x-1=2\left(x^2+5x-\frac{1}{2}\right)=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{27}{4}\right)=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\)
\(\Rightarrow Bmin=\frac{-27}{2}.''=''\Leftrightarrow x=\frac{-5}{2}\)
a)\(A=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu = khi \(x=\frac{-1}{2}\)
Vậy MinA=10 khi \(x=\frac{-1}{2}\)
b)\(B=3x^2-6x+1\)
\(=3x^2-6x+3-2\)
\(=3\left(x^2-2x+1\right)-2\)
\(=3\left(x-1\right)^2-2\ge-2\)
Dấu = khi \(x=1\)
Vậy MinB=-2 khi \(x=1\)
c)\(C=x^2-2x+y^2-4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy MinC=1 khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
a, Ta có: \(B=2x^2+10x-1=2x^2+10x+\dfrac{25}{2}-\dfrac{27}{2}\)
\(=2\left(x^2+2.x.\dfrac{5}{2}+\dfrac{25}{4}\right)-\dfrac{27}{2}\)
\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge\dfrac{-27}{2}\)
Dấu " = " khi \(2\left(x+\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{-5}{2}\)
Vậy \(MIN_B=\dfrac{-27}{2}\) khi \(x=\dfrac{-5}{2}\)
b, Ta có: \(C=5x-x^2=-\left(x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
Dấu " = " khi \(-\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(MAX_C=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)
\(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra <=> x = 3
Vậy MinA = 1
\(B=5x^2-10x+3=5\left(x^2-2x+1\right)-2=5\left(x-1\right)^2-2\ge-2\forall x\)
Dấu "=" xảy ra <=> x = 1
Vậy MinB = -2
\(C=2x^2+8x+y^2-10y+43=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)
Dấu "=" xảy ra <=> x = -2 ; y = 5
Vậy MinC = 10
\(A=x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)^2+1\ge1\forall x\)
Dấu"=" xảy ra khi \(x-3=0\Leftrightarrow x=3\)
Vậy \(Min_A=1\Leftrightarrow x=3\)
b,\(B=5x^2-10x+3\)
\(=5\left(x^2-2x+1\right)-2\)
\(=5\left(x-1\right)^2-2\ge-2\forall x\)
Dấu"=" xảy ra khi \(x-1=0\Leftrightarrow x=1\)
Vậy \(Min_B=-2\Leftrightarrow x=1\)
c,\(C=2x^3+8x+y^2-10+43\)
\(=2x^2+8x+8+y^2-10y+25+10\)
\(=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10\)
\(=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)
Dấu"=" xảy ra khi \(\orbr{\begin{cases}x+2=0\\y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\y=5\end{cases}}}\)
Vậy \(Min_C=10\Leftrightarrow x=-2;y=5\)
Ta có : A = x2 - 6x + 15
= x2 - 6x + 9 + 6
= (x - 3)2 + 6 \(\ge6\forall x\in R\)
Vậy Amin = 6 khi x = 3.
a) \(A=x^2-6x+11\)
\(\Rightarrow A=x^2-6x+9+2\)
\(\Rightarrow A=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = 3
Vậy \(MIN\) \(A=2\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(\Rightarrow B=2\left(x^2+5\right)-1\)
\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{25}{2}-1\)
\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\)
Ta có: \(2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)\ge0\forall x\)
\(\Rightarrow2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\ge-\dfrac{23}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{-5}{2}\)
Vậy \(MIN\) \(B=\dfrac{-23}{2}\Leftrightarrow x=\dfrac{-5}{2}\)
c) \(C=5x-x^2\)
\(\Rightarrow C=-\left(x^2-5x\right)\)
\(\Rightarrow C=-\left(x^2-2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(\Rightarrow C=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)
Ta có: \(-\left(x-\dfrac{5}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)
Vậy \(MAX\) \(C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)