\(1.25-\left|\frac{x}{3}+2\right|=\left(-\frac{1}{2}\right)^3\)
\(1.25-\left|\frac{x}{3}+2\right|=-\frac{1}{8}\)
\(\left|\frac{x}{3}+2\right|=1.25-\left(-\frac{1}{8}\right)\)
\(\left|\frac{x}{3}+2\right|=\frac{11}{8}\)
\(\Rightarrow\) \(\frac{x}{3}+2=\frac{11}{8}\) hay \(\frac{x}{3}+2=-\frac{11}{8}\)
TH1: \(\frac{x}{3}+2=\frac{11}{8}\)
\(\frac{x}{3}=\frac{11}{8}-2\)
\(\frac{x}{3}=\frac{-5}{8}\)
\(\Rightarrow x×8=3×\left(-5\right)\)
\(x×8=-15\)
\(x=-\frac{15}{8}\)
TH2: \(\frac{x}{3}+2=-\frac{11}{8}\)
\(\frac{x}{3}=-\frac{11}{8}-2\)
\(\frac{x}{3}=\frac{-27}{8}\)
\(\Rightarrow x×8=3×\left(-27\right)\)
\(x×8=-81\)
\(x=-\frac{81}{8}\)
Vậy \(x\in\left\{-\frac{5}{8};-\frac{81}{8}\right\}\)
Cbht
