Đặt:
\(A=2x^2-6x\)
\(A=2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\)
\(A=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\)
\(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) nên \(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" xảy ra khi:
\(x=-\dfrac{3}{2}\)
\(2x^2-6x\)
\(=2.\left(x^2-3x\right)\)
=\(2\left[x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3^{ }}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)
\(=2\left[\left(x-\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)
=\(2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\ge2\left(0-\dfrac{9}{4}\right)\ge0\)
Vậy GTNN của biểu thức là\(\dfrac{-9}{2}\) xẩy ra khi \(x=\dfrac{3}{2}\)
Nguồn: OLM
Bạn học tốt nhé!
\(2x^2-6x\\ =2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\\ =2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ =2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{2}\\ =2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
\(\text{Ta có: }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)
Dấu \("="\) xảy ra khi:
\(2\left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow \left(x-\dfrac{3}{2}\right)^2=0\\\Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(GTNN\) của biểu thức là \(-\dfrac{9}{2}\) khi \(x=\dfrac{3}{2}\)
