\(A=0,5-\left|3,4-x\right|\)
\(\left|3,4-x\right|\ge0\forall x\)
\(\Rightarrow0,5-\left|3,4-x\right|\le0,5\)
Dấu "=" xảy ra khi:
\(\left|3,4-x\right|=0\Rightarrow3,4-x=0\Rightarrow x=3,4\)
\(\Rightarrow MAX_A=0,5\) khi \(\) \(x=3,4\)
\(B=\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi:
\(\left|x-\dfrac{1}{2}\right|=0\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(\Rightarrow MIN_B=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
a, \(-\left|3,4-x\right|\le0\Rightarrow A=0,5-\left|3,4-x\right|\le0,5\)
Dấu " = " khi \(-\left|3,4-x\right|=0\Rightarrow x=3,4\)
Vậy \(MAX_A=0,5\) khi x = 3,4
b, \(\left|x-\dfrac{1}{2}\right|\ge0\Rightarrow B=\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu " = " khi \(\left|x-\dfrac{1}{2}\right|=0\Rightarrow x=\dfrac{1}{2}\)
Vậy \(MIN_B=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
A) \(A=0,5-\left|3,4-x\right|\)
Ta có: \(0,5-\left|3,4-x\right|\le0,5\)
Dấu "=" xảy ra khi \(\left|3,4-x\right|=0\Leftrightarrow x=3,4\)
Vậy \(Max_A=0,5\) khi \(x=3,4\)
B) \(B=\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}\)
Ta có: \(\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(\left|x-\dfrac{1}{2}\right|=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(Min_A=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)