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a, \(A=-x^2-2x+3=-\left(x^2+2x-3\right)=-\left(x^2+2x+1-4\right)\)
\(=-\left(x+1\right)^2+4\le4\)
Dấu ''='' xảy ra khi x = -1
Vậy GTLN là 4 khi x = -1
b, \(B=-4x^2+4x-3=-\left(4x^2-4x+3\right)=-\left(4x^2-4x+1+2\right)\)
\(=-\left(2x-1\right)^2-2\le-2\)
Dấu ''='' xảy ra khi x = 1/2
Vậy GTLN B là -2 khi x = 1/2
c, \(C=-x^2+6x-15=-\left(x^2-2x+15\right)=-\left(x^2-2x+1+14\right)\)
\(=-\left(x-1\right)^2-14\le-14\)
Vâỵ GTLN C là -14 khi x = 1
Bài 8 :
b, \(B=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)
Dấu ''='' xảy ra khi x = 3
Vậy GTNN B là 2 khi x = 3
c, \(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu ''='' xảy ra khi x = 1/2
Vậy ...
c, \(x^2-12x+2=x^2-12x+36-34=\left(x-6\right)^2-34\ge-34\)
Dấu ''='' xảy ra khi x = 6
Vậy ...
\(B1,a,A=x^2-6x+11\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\ge2\)
Dấu "=" <=> x=3
Vậy ..........
\(b,B=x^2-20x+101\)
\(=\left(x^2-20x+100\right)+1\)
\(=\left(x-10\right)^2+1\ge1\)
Dấu "=" <=> x = 10
Vậy .
\(2,a,A=4x-x^2+3\)
\(=7-\left(x^2-4x+4\right)\)'
\(=7-\left(x-2\right)^2\le7\)
Dấu ''='' <=> x = 2
Vậy .
\(b,B=-x^2+6x-11\)
\(=-2-\left(x^2-6x+9\right)\)
\(=-2-\left(x-3\right)^2\le-2\)
Dấu ""=" <=> x = 3
Vậy..
\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)
a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Dấu "=" \(\Leftrightarrow x=-1\)
b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)
c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)
Dấu "=" \(\Leftrightarrow x=2\)
\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
b:
\(D=-25x^2+10x-1-10\)
\(=-\left(25x^2-10x+1\right)-10\)
\(=-\left(5x-1\right)^2-10< =-10\)
Dấu = xảy ra khi x=1/5
\(E=-9x^2-6x-1+20\)
\(=-\left(9x^2+6x+1\right)+20\)
\(=-\left(3x+1\right)^2+20< =20\)
Dấu = xảy ra khi x=-1/3
\(F=-x^2+2x-1+1\)
\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)
`A=x^4-6x^3+18x^2-6xy+y^2+2012`
`=x^4-6x^3+9x^2+9x^2-6xy+y^2+2012`
`=(x^2-x)^2+(3x-y)^2+2012>=2012`
Dấu "=" xảy ra khi:
$\begin{cases}x=x^2\\y=3x\end{cases}$
`<=>` $\left[ \begin{array}{l}\begin{cases}x=0\\y=3x=0\\\end{cases}\\\begin{cases}x=1\\y=3x=3\\\end{cases}\end{array} \right.$
Vậy `min_A=2012<=>` $\left[ \begin{array}{l}x=y=0\\\begin{cases}x=1\\y=3\end{cases}\end{array} \right.$