\(A=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10=\left(x-1\right)\left(x-6\right)\left(x-3\right)\left(x-4\right)+10\)
\(=\left(x^2-7x+6\right)\left(x^2-7x+12\right)+10\)
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 =)))
\(\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\)
\(=\left(x-1\right)\left(x-6\right)\left(x-3\right)\left(x-4\right)+10\)
\(=\left(x^2-6x-x+6\right)\left(x^2-4x-3x+12\right)+10\)
\(=\left(x^2-7x+9-3\right)\left(x^2-7x+9+3\right)+10\)
\(=\left(x^2-7x+9\right)^2-9+10=\left(x^2-7x+9\right)^2+1\ge1\)
\("="\Leftrightarrow x^2-7x+9=0\)
\(\Leftrightarrow x^2-7x+\dfrac{49}{4}-\dfrac{13}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{2}+\sqrt{\dfrac{13}{4}}\right)\left(x-\dfrac{7}{2}-\sqrt{\dfrac{13}{4}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\sqrt{\dfrac{13}{4}}\\x=\dfrac{7}{2}+\sqrt{\dfrac{13}{4}}\end{matrix}\right.\)
