Question

tìm so tu nhien x biet (1/1.2.3 + 1/2.3.4+...+ 1/8.9.10).x=23/45

Answer 1

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}.\frac{44}{90}.x=\frac{23}{45}\Rightarrow\frac{11}{45}.x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)