| x + 1/2 | = 7/4
=> \(\hept{\begin{cases}x+\frac{1}{2}=\frac{7}{4}\\x+\frac{1}{2}=\frac{-7}{4}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{4}\\x=\frac{-9}{4}\end{cases}}}\)
\(3-\left|x+\frac{1}{2}\right|=\frac{5}{4}\)
\(3-x+\frac{1}{2}=\frac{5}{4}\)
\(x+\frac{1}{2}=3-\frac{5}{4}\)
\(x+\frac{1}{2}=\frac{7}{4}\)
\(x=\frac{7}{4}-\frac{1}{2}\)
\(x=\frac{5}{4}\)
\(3-\left|x+\frac{1}{2}\right|=\frac{7}{4}\)
\(\left|x+\frac{1}{2}\right|=\frac{-7}{4}\)
\(x+\frac{1}{2}=\frac{-7}{4}\)
\(x=\frac{-7}{4}-\frac{1}{2}\)
\(x=-\frac{9}{4}\)
\(3-!x+\frac{1}{2}!=\frac{5}{4}\)
\(!x+\frac{1}{2}!=3-\frac{5}{4}\)
\(!x+\frac{1}{2}!=\frac{12}{4}-\frac{5}{4}\)
\(!x+\frac{1}{2}!=\frac{7}{4}\)
\(TH1:x+\frac{1}{2}=\frac{7}{4}\Rightarrow x=\frac{5}{4}\)
\(TH2:x+\frac{1}{2}=\frac{-7}{4}\Rightarrow x=\frac{-9}{4}\)
Vậy \(x\in\left\{\frac{5}{4};\frac{-9}{4}\right\}\)
3-|x + 1/2| = 5/4
|x + 1/2| = 3 - 5/4
|x + 1/2| = 7/4
TH1 x + 1/2 = 7/4 TH2 x + 1/2 = -7/4
x = 7/4 -1/2 x = -7/4 -1/2
x = 5/4 x = -9/4
3 - /x + 1/2 / = 5/ 4
Th1 : 3 - / x + 1/2 / =-5/4
/ x +1/2 / = 3-(-5/4)
/ x + 1/2 / = 8,4
x + 1/2 = 8,4
x = 8,4 -1/2
x = 7,9
Vậy Th1 x = 7,9
Th2 : 3 - / x + 1/2 / = 5/4
/ x + 1/2 / = 3 - 5/4
/ x + 1/2 / = 7/4
x + 1/2 = 7/4
x + 1/2 = 7/4 - 1.2
x + 1/2 = 5/4
