\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
TH1: \(x+\frac{3}{4}-\frac{1}{3}=0\)
<=> x=\(-\frac{5}{12}\)
TH2: \(x+\frac{3}{4}+\frac{1}{3}=0\)
<=> x=\(-\frac{13}{12}\)
kl có 2 giá trị x
\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
\(\left|x+\frac{3}{4}\right|=0+\frac{1}{3}\)
\(x+\frac{3}{4}=\pm\frac{1}{3}\)
Nếu \(x+\frac{3}{4}=\frac{1}{3}\)\(x=\frac{1}{3}-\frac{3}{4}\)
\(x=-\frac{5}{12}\)
Nếu \(x+\frac{3}{4}=-\frac{1}{3}\)\(x=\left(-\frac{1}{3}\right)-\frac{3}{4}\)
\(x=-1\frac{1}{12}\)
Vậy \(x=-\frac{5}{12}\) hoặc \(x=-1\frac{1}{12}\)