Question

Answer 1

Lời giải:
Gọi tử số là $T$

\(T=(1-\frac{1}{6})+(1-\frac{2}{7})+(1-\frac{3}{8})+....+(1-\frac{88}{93})\)

\(=\frac{5}{6}+\frac{5}{7}+\frac{5}{8}+....+\frac{5}{93}=5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})\)

Gọi mẫu số là $M$
\(M=\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+....+\frac{1}{93})\)

Do đó:
\(C=\frac{5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}{\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}=\frac{5}{\frac{-1}{2}}=-10\)