\(P=\frac{3}{2}+\left|2x-5\right|\)
+Có: \(\left|2x-5\right|\ge0với\forall x\\ \Rightarrow\frac{3}{2}+\left|2x-5\right|\ge\frac{3}{2}\\ \Leftrightarrow P\ge\frac{3}{2}\)
+Dấu "=" xảy ra khi \(\left|2x-5\right|=0\Leftrightarrow x=\frac{5}{2}\)
+Vậy \(P_{min}=\frac{3}{2}\) khi \(x=\frac{5}{2}\)
Ta có:|2x-5| \(\ge\)0 với mọi x;cộng \(\frac{2}{3}\) vào hai vế.
\(\Rightarrow\)|2x-5| + \(\frac{2}{3}\) \(\ge\)0 +\(\frac{2}{3}\) với mọi x
\(\Rightarrow\)|2x-5| + \(\frac{2}{3}\) \(\ge\)\(\frac{2}{3}\) với mọi x
Vậy |2x-5| + \(\frac{2}{3}\) =\(\frac{2}{3}\) là Min khi 2x-5=0 \(\Rightarrow\)2x=5 \(\Rightarrow\)x=\(\frac{5}{2}\)
