Tìm x, biết 2/5+3/5.(3x-3,7)=-53/10

a.

\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{2}\)

\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)

\(-\frac{5}{6}\times x=\frac{5}{2}\)

\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)

\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)

\(x=-3\)

b.

\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)

\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)

\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)

\(3x-\frac{37}{10}=-\frac{19}{2}\)

\(3x=-\frac{19}{2}+\frac{37}{10}\)

\(3x=\frac{-95+37}{10}\)

\(3x=-\frac{58}{10}\)

\(3x=-\frac{29}{5}\)

\(x=-\frac{29}{5}\div3\)

\(x=-\frac{29}{5}\times\frac{1}{3}\)

\(x=-\frac{29}{15}\)

c.

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21-16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}\times\frac{4}{3}\)

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

d.

\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)

\(-\frac{2}{3}\times x=\frac{3}{10}-\frac{1}{5}\)

\(-\frac{2}{3}\times x=\frac{3-2}{10}\)

\(-\frac{2}{3}\times x=\frac{1}{10}\)

\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)

\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)

\(x=-\frac{3}{20}\)

e.

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

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

\(\left|x\right|=\frac{20+9}{12}\)

\(\left|x\right|=\frac{29}{12}\)

\(x=\pm\frac{29}{12}\)

Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)

f.

\(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)

\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)

\(\left|2x-\frac{1}{3}\right|=\frac{6-5}{6}\)

\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)

\(2x-\frac{1}{3}=\pm\frac{1}{6}\)

• \(2x-\frac{1}{3}=\frac{1}{6}\)

                \(2x=\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{1+2}{6}\)

                \(2x=\frac{3}{6}\)

                \(2x=\frac{1}{2}\)

                  \(x=\frac{1}{2}\div2\)

                  \(x=\frac{1}{2}\times\frac{1}{2}\)

                  \(x=\frac{1}{4}\)

• \(2x-\frac{1}{3}=-\frac{1}{6}\)

                \(2x=-\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{-1+2}{6}\)

                \(2x=\frac{1}{6}\)

                 \(x=\frac{1}{6}\div2\)

                 \(x=\frac{1}{6}\times\frac{1}{2}\)

                 \(x=\frac{1}{12}\)

Vậy x = 1/4 hoặc x = 1/12.

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