\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{45}{90}-\frac{1}{90}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{44}{90}.x=\frac{22}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{22}{45}\)
\(\Rightarrow x=\frac{22}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{22}{45}.\frac{45}{11}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Chúc học tốt !!!
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
VD ( dễ hiểu )
còn của bn là vd
\(\frac{1}{a\cdot b\cdot c}=\frac{1}{d}\left(\frac{1}{a\cdot b}-\frac{1}{b\cdot c}\right)\)
trong đó : d là khoảng cách
a,b,c là hằng số khác 0
(nếu nhớ ko nhầm)
