Tìm \(x\), biết:

    a) \(5,4:x =  - 1,2\);                                     b) \( - x - \frac{3}{2} = \frac{{ - 5}}{4}\);

    c) \(\frac{{ - 5}}{{ - 14}} = \frac{{20}}{{6 - 5x}}\);       d) \(\left( {3x - 1} \right)\left( { - \frac{1}{2}x + 5} \right) = 0\).

a) \(5,4:x =  - 1,2\)

    \(x = 5,4:\left( { - 1,2} \right)\)

    \(x =  - 4,5\).

Vậy \(x =  - 4,5\).

c) \(\frac{{ - 5}}{{ - 14}} = \frac{{20}}{{6 - 5x}}\)

Suy ra \( - 5.\left( {6 - 5x} \right) = \left( { - 14} \right).20\)

\(6 - 5x = 56\)

\(5x = 6 - 56\)

\(5x =  - 50\)

\(x =  - 10\)

Vậy \(x =  - 10\).

b) \( - x - \frac{3}{2} = \frac{{ - 5}}{4}\)

\( - x =  - \frac{5}{4} + \frac{3}{2}\)

\( - x = \frac{1}{4}\)

\(x =  - \frac{1}{4}\)

Vậy \(x =  - \frac{1}{4}\).

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

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:

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

\( - \frac{1}{2}x =  - 5\)

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

\(x = 10\).