\(\left(2x-\dfrac{3}{4}\right)^2=1\)
\(\left(2x-\dfrac{3}{4}\right)^2=1^2\)
\(\circledast\)TH1: \(2x-\dfrac{3}{4}=1\)
\(2x=1+\dfrac{3}{4}\)
\(2x=\dfrac{4}{4}+\dfrac{3}{4}\)
\(2x=\dfrac{7}{4}\)
\(x=\dfrac{7}{4}:2\)
\(x=\dfrac{7}{4}\cdot\dfrac{1}{2}\)
\(x=\dfrac{7}{8}\)
\(\circledast\)TH2: \(2x-\dfrac{3}{4}=-1\)
\(2x=-1+\dfrac{3}{4}\)
\(2x=-\dfrac{4}{4}+\dfrac{3}{4}\)
\(2x=-\dfrac{1}{4}\)
\(x=-\dfrac{1}{4}:2\)
\(x=-\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(x=-\dfrac{1}{8}\)
Vậy \(x\in\left\{\dfrac{7}{8};-\dfrac{1}{8}\right\}\).
mik sợ quá sợ ko đạt giải thi 3 môn luôn toán, văn, anh
