Tìm x:

a) $\frac{2}{3}+\frac{1}{3}\cdot x=\frac{5}{6}$;

b) 53,2 : (x – 3,5) + 45,8 = 99

c) \[\left( {4\frac{1}{2} - 2x} \right).1\frac{4}{{61}} = 6\frac{1}{2}\];

d) \(\frac{1}{2}\,\, \cdot \,x\,\, + \,\,150\% \cdot \,\,x\,\, = \,\,\,2022\)

Hướng dẫn giải:

a) \(\frac{2}{3}\,\, + \,\,\frac{1}{3} \cdot \,\,x\, = \,\frac{5}{6}\)

\({\mkern 1mu} {\mkern 1mu} \frac{1}{3}{\mkern 1mu} \cdot \,x{\mkern 1mu} \, = \,\,\frac{5}{6} - \frac{2}{3}{\mkern 1mu} \)

\({\mkern 1mu} \frac{1}{3}{\mkern 1mu} \, \cdot \,x{\mkern 1mu} \, = {\mkern 1mu} {\mkern 1mu} \frac{5}{6}{\mkern 1mu} \, - \,{\mkern 1mu} \frac{4}{6}\)

\({\mkern 1mu} \frac{1}{3} \cdot \,{\mkern 1mu} x{\mkern 1mu} \, = {\mkern 1mu} \,{\mkern 1mu} \frac{1}{6}{\mkern 1mu} \)

\(x{\mkern 1mu} {\mkern 1mu} = \,{\mkern 1mu} {\mkern 1mu} \frac{1}{6}\,{\mkern 1mu} {\mkern 1mu} :{\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} \frac{1}{3}\)

\(x{\mkern 1mu} \, = \,{\mkern 1mu} {\mkern 1mu} \frac{1}{6}\,{\mkern 1mu} \cdot \,\frac{3}{1}\)

\(x{\mkern 1mu} \, = \,{\mkern 1mu} \frac{1}{2}\).

Vậy \(x{\mkern 1mu} \, = \,{\mkern 1mu} \frac{1}{2}\).

b) 53,2 : (x – 3,5) + 45,8 = 99

53,2 : (x – 3,5) = 99 – 45,8

53,2 : (x – 3,5) = 53,2

x – 3,5 = 53,2 : 53,2

x – 3,5 = 1

x = 1 + 3,5

x = 4,5.

Vậy x = 4,5.

c) \[\left( {4\frac{1}{2} - 2x} \right).1\frac{4}{{61}} = 6\frac{1}{2}\].

\[\left( {\frac{9}{2} - 2x} \right).\frac{{65}}{{61}} = \frac{{13}}{2}\]

\[\frac{9}{2} - 2x = \frac{{13}}{2}:\frac{{65}}{{61}}\]

\[\frac{9}{2} - 2x = \frac{{13}}{2}.\frac{{61}}{{65}}\]

\[\frac{9}{2} - 2x = \frac{{13}}{2}.\frac{{61}}{{5.13}}\]

\[\frac{9}{2} - 2x = \frac{{61}}{{10}}\]

\[2x = \frac{9}{2} - \frac{{61}}{{10}}\]

\[2x = \frac{{45}}{{10}} - \frac{{61}}{{10}}\]

\[2x = \frac{{ - 16}}{{10}}\]

\[2x = \frac{{ - 8}}{5}\]

\[x = \frac{{ - 8}}{5}:2\]

\[x = \frac{{ - 8}}{5}.\frac{1}{2}\]

\[x = \frac{{ - 4}}{5}\]

Vậy \[x = \frac{{ - 4}}{5}\].

d) \(\frac{1}{2}\,\, \cdot \,x\,\, + \,\,150\% \cdot \,\,x\,\, = \,\,\,2022\)

\(\frac{1}{2}{\mkern 1mu} {\mkern 1mu} \cdot \,x{\mkern 1mu} \,{\mkern 1mu} + \,{\mkern 1mu} {\mkern 1mu} \frac{{150}}{{100}}\,\, \cdot \,{\mkern 1mu} x\,{\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2022\)

\(\frac{1}{2}{\mkern 1mu} {\mkern 1mu} \cdot \,{\mkern 1mu} x{\mkern 1mu} {\mkern 1mu} \, + \,{\mkern 1mu} {\mkern 1mu} \frac{3}{2}{\mkern 1mu} \, \cdot {\mkern 1mu} {\mkern 1mu} \,x{\mkern 1mu} \,{\mkern 1mu} = \,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2022\)

\(x.\left( {\frac{1}{2} + \frac{3}{2}} \right) = 2022\)

\(x\,.{\mkern 1mu} \,\frac{4}{2}{\mkern 1mu} \, = {\mkern 1mu} \,2022\)

x . 2 = 2022

x = 2022 : 2

x = 1011

Vậy x = 1011.