\(d)\)\(\left(2x-\frac{1}{4}\right)^2=\frac{49}{25}\)

\(\left(2x-\frac{1}{4}\right)^2=\left(\pm\frac{7}{5}\right)^2\)

\(\orbr{\begin{cases}2x-\frac{1}{4}=\frac{7}{5}\\2x-\frac{1}{4}=-\frac{7}{5}\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}2x=\frac{33}{20}\\2x=-\frac{23}{20}\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{33}{40}\\x=-\frac{23}{40}\end{cases}}\)

Vậy \(x\in\left\{-\frac{23}{40};\frac{33}{40}\right\}\)

- Dấu \(\pm\)là dấu cộng, trừ nha bn

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{4}=\frac{7}{5}\\2x-\frac{1}{4}=-\frac{7}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{33}{40}\\x=-\frac{23}{40}\end{cases}}\)

B  , 6,67

mik nghi the

\(A=\left|x+\frac{2}{3}\right|+1\frac{2}{3}\)

\(A=\left|x+\frac{2}{3}\right|+\frac{5}{3}\)

Áp dungk KT \(\left|x\right|\ge0\)\(\forall\)\(x\)

BG :

Ta có : \(\left|x+\frac{2}{3}\right|\ge0\)\(\forall\)\(x\)

\(\Leftrightarrow\)\(\left|x+\frac{2}{3}\right|+\frac{5}{3}\ge0+\frac{5}{3}\)\(\forall\)\(x\)

hay \(A\ge\frac{5}{3}\)\(\forall\)\(x\)

Dấu "=" xảy ra :

\(\Leftrightarrow\)\(\left|x+\frac{2}{3}\right|=0\)

\(\Leftrightarrow\)\(x+\frac{2}{3}=0\)

\(\Leftrightarrow\)\(x=-\frac{2}{3}\)

Vậy GTNN của \(A\) bằng \(\frac{5}{3}\)đạt được khi \(x=-\frac{2}{3}\)