a) \(\frac{1}{3} + \frac{2}{3}:x = - 7\)

\(\frac{2}{3}:x = - 7 - \frac{1}{3}\)

\(\frac{2}{3}:x = - \frac{{22}}{3}\)

\(x = \frac{2}{3}:\left( { - \frac{{22}}{3}} \right)\)

\(x = \frac{2}{3}.\left( { - \frac{3}{{22}}} \right)\)

b) \(\frac{{x + 1}}{2} = \frac{{x - 1}}{3}\)

\(3\left( {x + 1} \right) = 2\left( {x - 1} \right)\)

\(3x + 3 = 2x - 2\)

\(3x - 2x = - 2 - 3\)

\(x = - 5\)

Vậy \(x = - 5\).

c) \(\frac{1}{4} + \left| {3x - 1\frac{1}{4}} \right| = \frac{3}{2}\)

\(\left| {3x - \frac{5}{4}} \right| = \frac{3}{2} - \frac{1}{4}\)

\(\left| {3x - \frac{5}{4}} \right| = \frac{5}{4}\)

Trường hợp 1:

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

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

\(3x = \frac{5}{2}\)

\(x = \frac{5}{2}:3\)

\(x = \frac{5}{6}\)

Vậy \(x \in \left\{ {\frac{5}{6};0} \right\}\).

Trường hợp 2:

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

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

\(3x = 0\)

\(x = 0\)

Hai góc đối đỉnh thì A. Bù nhau; B. Kề nhau; C. Bằng nhau; D. Kề bù.