Question

[Nguyễn Chí Thành]

(\(\frac{1}{1.101}\)+    \(\frac{1}{2.102}\)+...+     \(\frac{1}{10.110}\)).x=\(\frac{1}{1.11}\)+    \(\frac{1}{2.12}\)+...+\(\frac{1}{100.110}\)

Tìm x

Answer 1

[KAl(SO4)2·12H2O]

\(\Rightarrow\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right).x=10.\left(\frac{10}{1.10}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right).x=10\)\(\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(\Rightarrow\left(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)- \left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right).x=10\)

\(\left(\left(1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right)\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right).x=10\)

\(\left(\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right)\Rightarrow x=10\)