Question

Bài 1 ; Tính \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.........\frac{899}{900}\) \(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\) Bài 2 ; So sánh a.\(\frac{-8}{9}và\frac{-7}{9}\) b.\(\frac{-6}{7}và\frac{-7}{8}\) c. \(\frac{-11}{12}và\frac{17}{18}\) 

Answer 1

A. \(\frac{3}{4}\) x \(\frac{8}{9}\)x \(\frac{15}{16}\)x .... x \(\frac{899}{900}\)

= \(\frac{1.3}{2^2}\) x \(\frac{2.4}{3^3}\)x \(\frac{3.5}{4^2}\)x ... x \(\frac{29.31}{30^2}\)

= \(\left(\frac{1.2.3...29}{2.3.4...30}\right).\left(\frac{3.4.5...31}{2.3.4...30}\right)\)

= \(\frac{1}{30}.\frac{31}{2}\)= \(\frac{31}{60}\)

B. 

\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{8}{24}+\frac{9}{24}-\frac{14}{24}=\frac{8+9-14}{24}=\frac{3}{24}=\frac{1}{8}\)