Question

[Chibi Yoona]

Tính GTNN của biểu thức:

c) C= (x² - 3x + 1) (x² - 3x + 1)

d) D= (x² - 4x + 1) (x² - 4x +5)

Answer 1

[Vũ Hạ Linh]

a)C=(x²-3x+1)²>=0

Answer 2

[Khôi Bùi]

c ) \(C=\left(x^2-3x+1\right)\left(x^2-3x+1\right)=\left(x^2-3x+1\right)^2\ge0\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x^2-3x+1=0\)

\(\Leftrightarrow x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy Min C là : \(0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)

d ) \(D=\left(x^2-4x+1\right)\left(x^2-4x+5\right)\)

\(=\left(x^2-4x+3-2\right)\left(x^2-4x+3+2\right)\)

\(=\left(x^2-4x+3\right)^2-4\ge-4\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy Min D là : \(-4\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)