Question

Tìm x để biểu thức:

B= \(\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\) đạt giá trị lớn nhất

Answer 1

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)

Vì \(\left|2x+\dfrac{2}{3}\right|\ge0\Rightarrow\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)

=> Max_B=2/3 => 2x+2/3=0 <=> x=-1/3

Vậy Max_B=2/3 khi x=-1/3

Answer 2

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)

\(\text{Ta có : }\left|2x+\dfrac{2}{3}\right|\ge0\text{ }\forall\text{ }x\\ \Rightarrow B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)

\(\text{Dấu "=" xảy ra khi : }\left|2x+\dfrac{2}{3}\right|=0\\ \Leftrightarrow2x+\dfrac{2}{3}=0\\ \Leftrightarrow2x=-\dfrac{2}{3}\\ \Leftrightarrow x=-\dfrac{1}{3}\)

Vậy \(x=-\dfrac{1}{3}\)

Answer 3

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)

\(\left|2x+\dfrac{2}{3}\right|\ge0\)

\(B_{MAX}\Rightarrow\left|2x+\dfrac{2}{3}\right|_{MIN}\)

\(\left|2x+\dfrac{2}{3}\right|_{MIN}=0\)

\(\Rightarrow B_{MAX}=\dfrac{2}{3}-0=\dfrac{2}{3}\)

Answer 4

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