Question

Tìm x để biểu thức :

a) A - 0,6 + \(\left|\dfrac{1}{2}-x\right|\)đạt giá trị nhỏ nhất.

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

Answer 1

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 \(\dfrac{2}{3}\Leftrightarrow x=-\dfrac{1}{3}\)

Answer 2

\(A=0,6+\left|\dfrac{1}{2}-x\right|\)

\(\left|\dfrac{1}{2}-x\right|\ge0\forall x\in R\)

\(A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\)

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:

\(\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow2x=-\dfrac{2}{3}\Leftrightarrow x=-\dfrac{1}{3}\)