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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=2\)
b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)
c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)
\(=2\div\frac{1}{2}=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}=\frac{3}{2}\)
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=1+1\)
\(=2\)
b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)
\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)
\(=\frac{25}{81}.\frac{3}{4}\)
\(=\frac{25}{108}\)
c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)
\(=\frac{7}{8}.2+\frac{9}{8}.2\)
\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)
\(=2.\frac{16}{8}\)
\(=2.2\)
\(=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,(11/15+4/15)+(5/7+2/7)
=1+1
=2
b,5/9x(1/2+6/4)
=5/9x2
=10/9
c,1/2:(7/8+9/8)
=1/2:2
=1
d,(17/10-7/10)+1/2
=1+1/2
=3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+.....+\frac{1}{14\times15}=\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{6\times5}+...+\frac{15-14}{14\times15}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
Do đó: \(\frac{4}{15}=\frac{x}{30}\)
\(x=\frac{4}{15}\times30=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{14\times15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{4}{15}=\frac{x}{30}\)
\(\Rightarrow4\times30=15\times x\)
\(120=15\times x\)
\(x=120\div15\)
\(x=8.\)
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{5}{15}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{4}{15}=\frac{x}{30}\)
\(\frac{8}{30}=\frac{x}{30}\)
\(x=8\)
hok tốt!!
Dấu "." là dấu nhân nhé
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2016.2018}\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2016.2018}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2017}{2018}\)
\(=\frac{2017}{4036}\)
\(x.\frac{2}{7}.\frac{3}{4}=\frac{5}{21}\)
\(\frac{3}{14}x=\frac{5}{21}\)
\(x=\frac{5}{21}:\frac{3}{14}\)
\(x=\frac{5}{21}.\frac{14}{3}\)
\(x=\frac{10}{9}\)
Vậy \(x=\frac{10}{9}\)
những câu bài 2 này là tìm x bình thường