Tìm x, biết:
a) \(x + 4,5 = 9,5\);
b) \(\frac{7}{5}x - \frac{1}{2} = \frac{3}{8}\);
c) \(\left| {3x - 1} \right| + \frac{1}{3} = \frac{1}{2}\)
a, b) Sử dụng quy tắc chuyển vế và thực hiện phép tính để tìm x.
c) Chuyển vế, sử dụng kiến thức \(\left| A \right| = k > 0\) thì xảy ra hai trường hợp: \(A = k\) hoặc \(A = - k\).
a) \(x + 4,5 = 9,5\)
\(x = 9,5 - 4,5\)
\(x = 5\)
Vậy \(x = 5\).
b) \(\frac{7}{5}x - \frac{1}{2} = \frac{3}{8}\)
\(\begin{array}{l}\frac{7}{5}x = \frac{3}{8} + \frac{1}{2}\\\frac{7}{5}x = \frac{7}{8}\\x = \frac{7}{8}:\frac{7}{5}\\x = \frac{7}{8}.\frac{5}{7}\\x = \frac{5}{8}\end{array}\)
Vậy \(x = \frac{5}{8}\).
c) \(\left| {3x - 1} \right| + \frac{1}{3} = \frac{1}{2}\)
\(\begin{array}{l}\left| {3x - 1} \right| = \frac{1}{2} - \frac{1}{3}\\\left| {3x - 1} \right| = \frac{1}{6}\end{array}\)
Suy ra \(3x - 1 = \frac{1}{6}\) hoặc \(3x - 1 = \frac{{ - 1}}{6}\)
TH1: \(3x - 1 = \frac{1}{6}\)
\(\begin{array}{l}3x = \frac{1}{6} + 1\\3x = \frac{7}{6}\\x = \frac{7}{6}:3\\x = \frac{7}{{18}}\end{array}\)
TH2: \(3x - 1 = \frac{{ - 1}}{6}\)
\(\begin{array}{l}3x = - \frac{1}{6} + 1\\3x = \frac{5}{6}\\x = \frac{5}{6}:3\\x = \frac{5}{{18}}\end{array}\)
Vậy \(x = \frac{7}{{18}};x = \frac{5}{{18}}\).