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\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,Ta có :\(A=x\left(x-6\right)=x^2-6x\)
\(=x^2-6x+9-9\)
\(=\left(x-3\right)^2-9\)
Vì: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\)\(\left(x-3\right)^2-9\ge-9\forall x\)
Hay: \(A\ge-9\forall x\)
Dấu = xảy ra khi (x-3)^2=0
<=>x=3
Vậy Min A= -9 tại x=3
b,Ta có: \(B=-3x\left(x+3\right)-7\)
\(=-3x^2-9x-7\)
\(=-3\left(x^2+3x+\frac{7}{3}\right)\)
\(=-3\left[\left(x^2+3x+\frac{9}{4}\right)+\frac{1}{12}\right]\)
\(=-3\left[\left(x+\frac{3}{2}\right)^2+\frac{1}{12}\right]\)
\(=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\)
Vì: \(-3\left(x+\frac{3}{2}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le\frac{-1}{4}\forall x\)
Hay \(B\le\frac{-1}{4}\forall x\)
Dấu = xảy ra khi \(-3\left(x+\frac{3}{2}\right)^2=0\)
\(\Rightarrow x=\frac{-3}{2}\)
Vậy Max B=-1/4 tại x=-3/2
a) \(A=x\left(x-6\right)=x^2-6x+9-9=\left(x-3\right)^2-9\ge-9\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=3\)
Vậy Min A = -9 khi x = 3
b) \(B=-3x\left(x+3\right)-7=-3x^2-9x-7=-3\left(x^2+9x+20,25\right)+53,75\)
\(=-3\left(x+4,5\right)^2+53,75\le53,75\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-4,5\)
Vậy Max B = 53,75 khi x = -4,5
![](https://rs.olm.vn/images/avt/0.png?1311)
Với \(x\ge\dfrac{1}{6}\Leftrightarrow A=5x^2-6x+1-1=5x^2-6x\)
\(A=5\left(x^2-2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{9}{5}=5\left(x-\dfrac{3}{5}\right)^2-\dfrac{9}{5}\ge-\dfrac{9}{5}\\ A_{min}=-\dfrac{9}{5}\Leftrightarrow x=\dfrac{3}{5}\left(1\right)\)
Với \(x< \dfrac{1}{6}\Leftrightarrow A=5x^2+6x-1-1=5x^2+6x-2\)
\(A=5\left(x^2+2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{19}{5}=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{19}{5}\ge-\dfrac{19}{5}\\ A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\)
Với \(x\ge\dfrac{1}{3}\Leftrightarrow B=9x^2-6x-4\left(3x-1\right)+6=9x^2-18x+10\)
\(B=9\left(x^2-2x+1\right)+1=9\left(x-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=1\left(1\right)\)
Với \(x< \dfrac{1}{3}\Leftrightarrow B=9x^2-6x+4\left(3x-1\right)+6=9x^2+6x+2\)
\(B=\left(9x^2+6x+1\right)+1=\left(3x+1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=-\dfrac{1}{3}\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrow B_{min}=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-9x^2+26x-24\)
\(=x^3-4x^2-5x^2+20x+6x-24\)
\(=\left(x-4\right)\left(x^2-5x+6\right)\)
\(=\left(x-4\right)\left(x-2\right)\left(x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10=\left(x-1\right)\left(x-6\right)\left(x-3\right)\left(x-4\right)+10\)
\(=\left(x^2-7x+6\right)\left(x^2-7x+12\right)+10\)
Đặt \(x^2-7x+6=a\Rightarrow A=a\left(a+6\right)+10=a^2+6a+10\)
\(A=a^2+6a+9+1=\left(a+3\right)^2+1\ge1\)
\(\Rightarrow A_{min}=1\) khi \(a=-3\Leftrightarrow x^2-7x+9=0\Rightarrow x=...\) (nghiệm xấu)
Vậy giá trị nhỏ nhất của đa thức đã cho là 1
Ta có:
\(\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10=\left(x^2-7x+6\right)\left(x^2-7x+12\right)+10\)
Đặt \(t=x^2-6x+6\) ta được:
\(t\left(t+6\right)+10=t^2+6t+10=\left(t+3\right)^2+1\)
Vì \(\left(t+3\right)^2\ge0\forall x\Rightarrow\left(t+3\right)^2+1\ge1\forall x\)
\(\Rightarrow Min=1\Leftrightarrow t=-3\)
\(\Rightarrow x^2-7x+9=-3\)
Giải nốt đi bạn =)))