Question

\(\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)

Answer 1

Lời giải:
Gọi biểu thức là $P$

\(P=\frac{1-4}{4}.\frac{1-9}{9}...\frac{1-100}{100}=\frac{(1-2)(1+2)(1-3)(1+3)...(1-10)(1+10)}{2^2.3^2....10^2}\)

\(=\frac{(1-2)(1-3)...(1-10)}{2.3....10}.\frac{(1+2)(1+3)...(1+10)}{2.3.4...10}\)

\(=\frac{-1.2.3...9}{2.3...10}.\frac{3.4....11}{2.3.4..10}=\frac{-1}{10}.\frac{11}{2}=\frac{-11}{20}\)