Tìm min \(x^4+\left(x-3\right)^2\)
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a) A = (2x−1)(x−3)
=\(2x^2-6x-x+3=\left(2x^2-\frac{2.\sqrt{2}x.7}{2\sqrt{2}}+\frac{49}{8}\right)-\frac{49}{8}+3\)
=\(\left(\sqrt{2}x-\frac{7}{2\sqrt{2}}\right)^2-\frac{25}{8}\)>=\(-\frac{25}{8}\)
dấu = xảy ra khi x=\(\frac{7}{4}\)
=> Min A=\(-\frac{25}{8}\) khi x=7/4
b) B = (1−2x)(x−3)
=\(x-3+6x-2x^2=-\left(2x^2-7x+3\right)\)
=\(-\left(\sqrt{2}x-\frac{7}{2\sqrt{2}}\right)^2\)+\(\frac{49}{8}-3\)<=25/8
dấu = xảy ra khi x=7/4
=> Max B =25/8 khi x=7/4
Nhận thấy: |x-2017| = |-x+2017|
Áp dụng BĐT: |a| + |b| \(\ge\) |a+b|
=> A = |x-2016| + |-x+2017| \(\ge\) |x-2016+-x+2017| = |1| = 1
Vậy MinA = 1 khi \(2016\le x\le2017\)
\(A=\left|x-2016\right|+\left|x-2017\right|\)
Ta có : \(\begin{cases}\left|x-2016\right|\ge0\\\left|x-2017\right|\ge0\end{cases}\)
\(\Rightarrow\left|x-2016\right|+\left|x-2017\right|\ge0\)
\(\Rightarrow A\ge0\)
Dấu " = " xảy ra khi và chỉ khi \(\begin{cases}x-2016=0\\x-2017=0\end{cases}\Leftrightarrow\begin{cases}x=2016\\x=2017\end{cases}\)
Vậy \(Min_A=0\Leftrightarrow\begin{cases}x=2016\\x=2017\end{cases}.}\)
\(y=\left|x-1\right|+\left|x-3\right|\)
x x-3 3 0 0 _ _ _ + + + 1 x-1
+) Với \(x< 1\Rightarrow y=1-x+3-x=4-2x\)
+) Với \(1\le x\le3\Rightarrow y=x-1+3-x=2\)
+) Với \(x>3\Rightarrow y=x-1+x-3=2x-4\)
y=/x-1/+/x-3/ 1 2 3 4 5 1 2 3 4 -1 -2 -3 -1 -2
b) \(\Rightarrow y_{Min}=3\) khi \(1\le x\le3\)