Tìm x, biết: a) 1,2x + 1/2 = 0,6

a) \(1,2x + \frac{1}{2} = 0,6\)

\(1,2x + 0,5 = 0,6\)

\(1,2x = 0,1\)

\(x = \frac{1}{{12}}\).

Vậy \(x = \frac{1}{{12}}\).

b) \(\left( {3x - 1} \right)\left( { - \frac{1}{2}x + 5} \right) = 0\)

Suy ra \(3x - 1 = 0\) hoặc \( - \frac{1}{2}x + 5 = 0\)

Trường hợp 1:

\(3x - 1 = 0\)

\(3x = 1\)

\(x = \frac{1}{3}\).

Vậy \(x \in \left\{ {\frac{1}{3};10} \right\}\).

Trường hợp 2:

\( - \frac{1}{2}x + 5 = 0\)

\( - \frac{1}{2}x =  - 5\)

\(x =  - 5:\left( {\frac{{ - 1}}{2}} \right)\)

\(x = 10\).