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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{48}{98}\)
\(A=\frac{8}{49}\)
A = \(\frac{1}{3}\).{ \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)}
A = \(\frac{1}{3}\).{\(\frac{1}{2}-\frac{1}{98}\)}
A = \(\frac{1}{3}.\left\{\frac{49}{98}-\frac{1}{98}\right\}\)
A=\(\frac{1}{3}.\frac{24}{49}\)
A = \(\frac{49}{98}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3x\left(x-\frac{2}{3}\right)=0
\)
\(\)TH1:
3x=0
x=0:3
x=0
TH2
\(x-\frac{2}{3}=0
\)
\(x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy x={0;\(\frac{2}{3}\)}
34 +14 :x=25
\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(x=\frac{1}{4}:-\frac{7}{20}\)
\(x=-\frac{20}{28}\)
\(x=-\frac{5}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\frac{1}{5}-x=1\frac{3}{5}+\frac{7}{10}\)
\(\frac{16}{5}-x=\frac{8}{5}+\frac{7}{10}\)
\(\frac{16}{5}-x=\frac{23}{10}\)
\(x=\frac{23}{10}-\frac{16}{5}\)
\(x=-\frac{9}{10}\)