Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) \(A=0,6+\left|\dfrac{1}{2}-x\right|\)
Ta có : \(\left|\dfrac{1}{2}-x\right|\ge0\)
\(\Leftrightarrow0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\)
Vậy GTNN là 0,6 khi \(x=\dfrac{1}{2}.\)
- Đề ghi ko hiểu ?
b ) \(\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)
Ta có : \(\left|2x+\dfrac{2}{3}\right|\ge0\)
\(\Leftrightarrow\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)
Vậy GTNN là \(\dfrac{2}{3}\Leftrightarrow x=-\dfrac{1}{3}\)
\(A=0,6+\left|\dfrac{1}{2}-x\right|\)
\(\left|\dfrac{1}{2}-x\right|\ge0\forall x\in R\)
\(A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\)
Dấu "=" xảy ra khi:
\(\left|\dfrac{1}{2}-x\right|=0\Rightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)
\(\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)
Dấu "=" xảy ra khi:
\(\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow2x=-\dfrac{2}{3}\Leftrightarrow x=-\dfrac{1}{3}\)
\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)
2.
a/\(A=5-I2x-1I\)
Ta thấy: \(I2x-1I\ge0,\forall x\)
nên\(5-I2x-1I\le5\)
\(A=5\)
\(\Leftrightarrow5-I2x-1I=5\)
\(\Leftrightarrow I2x-1I=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của \(A=5\Leftrightarrow x=\frac{1}{2}\)
b/\(B=\frac{1}{Ix-2I+3}\)
Ta thấy : \(Ix-2I\ge0,\forall x\)
nên \(Ix-2I+3\ge3,\forall x\)
\(\Rightarrow B=\frac{1}{Ix-2I+3}\le\frac{1}{3}\)
\(B=\frac{1}{3}\)
\(\Leftrightarrow B=\frac{1}{Ix-2I+3}=\frac{1}{3}\)
\(\Leftrightarrow Ix-2I+3=3\)
\(\Leftrightarrow Ix-2I=0\)
\(\Leftrightarrow x=2\)
Vậy GTLN của\(A=\frac{1}{3}\Leftrightarrow x=2\)
1/ \(A=3\left|2x-1\right|-5\)
Ta có: \(\left|2x-1\right|\ge0\)
\(\Rightarrow3\left|2x-1\right|\ge0\)
\(\Rightarrow3\left|2x-1\right|-5\ge-5\)
Để A nhỏ nhất thì \(3\left|2x-1\right|-5\)nhỏ nhất
Vậy \(Min_A=-5\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)
Vì \(\left|2x+\dfrac{2}{3}\right|\ge0\Rightarrow\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)
=> MaxB=2/3 => 2x+2/3=0 <=> x=-1/3
Vậy MaxB=2/3 khi x=-1/3
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)
\(\text{Ta có : }\left|2x+\dfrac{2}{3}\right|\ge0\text{ }\forall\text{ }x\\ \Rightarrow B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)
\(\text{Dấu "=" xảy ra khi : }\left|2x+\dfrac{2}{3}\right|=0\\ \Leftrightarrow2x+\dfrac{2}{3}=0\\ \Leftrightarrow2x=-\dfrac{2}{3}\\ \Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)
\(\left|2x+\dfrac{2}{3}\right|\ge0\)
\(B_{MAX}\Rightarrow\left|2x+\dfrac{2}{3}\right|_{MIN}\)
\(\left|2x+\dfrac{2}{3}\right|_{MIN}=0\)
\(\Rightarrow B_{MAX}=\dfrac{2}{3}-0=\dfrac{2}{3}\)
Ta có : B= \(\dfrac{2}{3}\) - \(\left|2x+\dfrac{2}{3}\right|\)
\(\left|2x+\dfrac{2}{3}\right|\) \(\ge\)0 \(\forall\) x
=> \(\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)\(\le\) \(\dfrac{2}{3}\)
- Dấu "=" xảy ra khi 2x + \(\dfrac{2}{3}\) = 0
=> 2x + \(\dfrac{2}{3}\)=0
2x = \(\dfrac{-2}{3}\)
=> x = \(\dfrac{-1}{3}\)
Vậy khi x = \(\dfrac{-1}{3}\)thì B đạt giá trị lớn nhất