by Tìm \(x,y\) biết:
\(\,\left| {2x – 1} \right| – \frac{1}{2} = \frac{1}{3}\)
A. hoặc \(x = \frac{11}{{12}}\).
B. \(x = \frac{1}{{12}}\).
C. \(x = \frac{{11}}{{12}}\)
D. \(x = \frac{{11}}{{12}}\) hoặc \(x = \frac{1}{{12}}\).
Hướng dẫn
Chọn đáp án là: D
Phương pháp giải:
Sử dụng \(\left| A \right| = B > 0\) thì \(A = B\) hoặc \(A = – B\).
\(\,\left| {2x – 1} \right| – \frac{1}{2} = \frac{1}{3}\)
\(\begin{array}{l}\left| {2x – 1} \right| = \frac{1}{3} + \frac{1}{2}\\\left| {2x – 1} \right| = \frac{5}{6}\end{array}\)
+) TH1: \(2x – 1 = \frac{5}{6}\)
\(\begin{array}{l}2x = \frac{5}{6} + 1\\2x = \frac{{11}}{6}\\x = \frac{{11}}{6}:2\\x = \frac{{11}}{{12}}\end{array}\)
+) TH2: \(2x – 1 = – \frac{5}{6}\)
\(\begin{array}{l}2x = – \frac{5}{6} + 1\\2x = \frac{1}{6}\\x = \frac{1}{6}:2\\x = \frac{1}{{12}}\end{array}\)
Vậy \(x = \frac{{11}}{{12}}\) hoặc \(x = \frac{1}{{12}}\).
Chọn D
