\(A=x^2-6x+3\)
\(=\left(x^2-6x+9\right)-6\)
\(=\left(x+3\right)^2-6\)
ma \(\left(x+3\right)^2\ge0\Leftrightarrow\left(x+3\right)^2-6\ge-6\)
vậy gtnn của A là -6 tại x=-3
\(B=x^2+3x+7=\left(x^2+2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{17}{4}\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\)
vay .............................................
2/
\(A=-x^2+4x+8=-\left(x^2-4x+4\right)+12=-\left(x-2\right)^2+12\le12\)
vay .........................................
\(B=-x^2+3x-5=-\left(x^2-2\frac{3}{2}x+\frac{9}{4}\right)-\frac{11}{4}=\left(x-\frac{3}{2}\right)^2-\frac{11}{4}\le-\frac{11}{4}\)
vay.....................................
nếu có sai mong bạn thông cảm
1/ Ta có: A\(=x^2-6x+3\)
\(=x^2-2.x.3+3^2-6\)
\(=\left(x-3\right)^2-6\ge-6\left(\forall x\right)\)
Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Rightarrow x=3\)
Vậy Min A = -6 khi x = 3.
Ta có: B = \(x^2+3x+7\)
\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{19}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min B = 19/4 khi x = 3/2.
2/
Ta có: \(A=-x^2+4x+8\)
\(=-\left(x^2-4x-8\right)=-\left(x^2-4x+4-12\right)\)
\(=-\left[\left(x-2\right)^2-12\right]\)
\(=-\left(x-2\right)^2+12\le12\left(\forall x\right)\)
Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Rightarrow x=2\)
Vậy Max A = 12 khi x =2.
Ta có: \(B=-x^2+3x-5\)
\(=-\left(x^2-3x+5\right)=-\left(x^2-3x+\frac{9}{4}+\frac{11}{4}\right)\)
\(=-\left[\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\right]=-\left(x-\frac{3}{2}\right)^2-\frac{11}{4}\le-\frac{11}{4}\left(\forall x\right)\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min B = -11/4 khi x =3/2.
Chúc bn hc tốt!
bài 1 câu a là 19/4 chứ k phải 17/4 nha bn
Bài 1)
\(A=x^2-6x+3=\left(x^2-2.x.3+9\right)-6\)
\(=\left(x-3\right)^2-6\ge-6\)
Vậy \(min_A=-6\),dấu ''='' xảy ra khi \(x-3=0\Leftrightarrow x=3\)
\(B=x^2+3x+7=\left(x^2+2.\frac{3}{2}.x+\frac{9}{4}\right)+\frac{19}{4}\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\)
Vậy \(min_B=\frac{19}{4}\)dấu bằng xảy ra khi \(x+\frac{3}{2}=0\Leftrightarrow x=\frac{-3}{2}\)
Bài 2)
\(A=-x^2+4x+8=-\left(x^2-2.x.2+4\right)+12\)
\(=12-\left(x-2\right)^2\le12\)
Vậy \(max_A=12\)dấu ''=''\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(B=-x^2+3x-5=-\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)-\frac{11}{4}\)
\(=\frac{-11}{4}-\left(x-\frac{3}{2}\right)^2\le\frac{-11}{4}\)
Vậy \(max_B=\frac{-11}{4}\)dấu ''=''\(\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Xong rồi đấy,chúc bạn học tốt
P/s:Nếu đúng thì bạn k cho mình vs nhé
