Question

Tim gia tri nho nhat cua cac bieu thuc sau:

a.(2x+1/3)^2-5/6

b.|2x-3|+|y-1/2|+3/4

Answer 1

a)Ta thấy:

\(\left(2x+\frac{1}{3}\right)^2\ge0\)

\(\Rightarrow\left(2x+\frac{1}{3}\right)^2-\frac{5}{6}\ge0-\frac{5}{6}=-\frac{5}{6}\)

\(\Rightarrow A\ge-\frac{5}{6}\)

Dấu "=" <=>x=-1/6

Vậy MinA=-5/6<=>x=-1/6

b)Ta thấy:\(\hept{\begin{cases}\left|2x+3\right|\\\left|y-\frac{1}{2}\right|\end{cases}\ge}0\)

\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|\ge0\)

\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)

\(\Rightarrow B\ge\frac{3}{4}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|2x-3\right|=0\\\left|y-\frac{1}{2}\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)

Vậy...