Ta có: \(A\left(x\right)=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\)
\(=\left[\left(x-1\right)\left(x-6\right)\right]\left[\left(x-3\right)\left(x-4\right)\right]+10\)
\(=\left(x^2-7x+6\right)\left(x^2-7x+12\right)+10\)
Đặt \(x^2-7x+9=t\)
Do đó: \(A=\left(t-3\right)\left(t+3\right)+10\)
\(=t^2-9+10=t^2+1\)
Ta có: \(t^2\ge0\forall x\)
\(\Rightarrow t^2+1\ge1\forall x\)
Dấu '=' xảy ra khi
\(t^2=0\)
⇔t=0
hay \(x^2-7x+9=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{4}-\frac{49}{4}+9=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2=\frac{13}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{7}{2}=\sqrt{\frac{13}{4}}\\x-\frac{7}{2}=-\sqrt{\frac{13}{4}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\frac{13}{4}}+\frac{7}{2}\\x=-\sqrt{\frac{13}{4}}+\frac{7}{2}\end{matrix}\right.\)
Vậy: Giá trị nhỏ nhất của biểu thức \(A\left(x\right)=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\) là 1 khi \(x\in\left\{\sqrt{\frac{13}{4}}+\frac{7}{2};-\sqrt{\frac{13}{4}}+\frac{7}{2}\right\}\)
A(x) = (x - 1)(x - 3)(x - 4)(x - 6) + 10
A(x) = (x2 - 7x + 6)(x2 - 7x + 12) + 10
Đặt a = x2 - 7x ta được:
A(x) = (a + 12)(a + 6) + 10 = a2 + 18a + 82 = (a + 9)2 + 1
Vì (a + 9)2 ≥ 0
Do đó A(x) ≥ 1
Dấu bằng xảy ra khi x2 - 7x = -9
⇔ x2 - 2.\(\frac{7}{2}\)x + \(\frac{49}{4}\) - \(\frac{-13}{4}\) = 0
⇔ (x - \(\frac{7}{2}\))2 - \(\frac{13}{4}\) = 0
⇔ (x - \(\frac{7+\sqrt{13}}{2}\))(x - \(\frac{7-\sqrt{13}}{2}\)) = 0
⇔x = \(\frac{7\pm\sqrt{13}}{2}\)
Vậy Min A(x) = 1 tại x = \(\frac{7\pm\sqrt{13}}{2}\)
Min = GTNN
