\(||3x-1|-\dfrac{1}{2}|=\dfrac{5}{2}\)
Có thể xảy ra 2 trường hợp:
TH1:\(||3x-1|-\dfrac{1}{2}|=-\dfrac{5}{2}\)
TH2: \(||3x-1|-\dfrac{1}{2}|=\dfrac{5}{2}\)
Giả sử \(|3x-1|-\dfrac{1}{2}=-\dfrac{5}{2}\)
⇔ \(|3x-1|=-\dfrac{5}{2}+\dfrac{1}{2}\)
⇔ \(|3x-1|=-2\) (Vô lí, vì |3x - 1| ≥ 0 ∀ x)
⇒ \(|3x-1|-\dfrac{1}{2}=\dfrac{5}{2}\)
⇔ \(|3x-1|=\dfrac{5}{2}+\dfrac{1}{2}\)
⇔ \(|3x-1|=3\)
⇔ \(3x-1\in\left\{\pm3\right\}\)
⇔ \(3x\in\left\{-2;4\right\}\)
⇔ \(x\in\left\{-\dfrac{2}{3};\dfrac{4}{3}\right\}\)
Vậy \(x\in\left\{-\dfrac{2}{3};\dfrac{4}{3}\right\}\)
\(\left|\left|3x-1\right|-\dfrac{1}{2}\right|=\dfrac{5}{2}\)
\(\Rightarrow\)2 trường hợp:
Th1:\(3x-1-\dfrac{1}{2}=\dfrac{5}{2}\)
\(3x-1=\dfrac{5}{2}+\dfrac{1}{2}\)
\(3x-1=3\)
\(3x=3+1\)
\(3x=4\Rightarrow x=4:3\Rightarrow x=\dfrac{4}{3}\)
Th2:
\(3x-1-\dfrac{1}{2}=-\dfrac{5}{2}\)
\(3x-1=-\dfrac{5}{2}+\dfrac{1}{2}\)
\(3x-1=-2\)
\(3x=-2+1\)
\(3x=-1\Rightarrow x=-1:3\Rightarrow x=\dfrac{-1}{3}\)
P/s Mình làm theo cách chửa mình nếu sai thì xin lỗi bạn nha
