\(\dfrac{2}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}...\dfrac{99}{10^2}\)
=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}\)
=\(\dfrac{1.2.3...9}{2.3.4...10}.\dfrac{3.4.5...11}{2.3.4...10}\)
Rút gọn biểu thức, ta được:
=\(\dfrac{1}{10}\).\(\dfrac{11}{2}\)
=\(\dfrac{11}{22}\)
=\(\dfrac{1}{2}\)
Đúng 0
Bình luận (3)
\(goiA=\dfrac{3}{2^2}.\dfrac{8}{3^2}......\dfrac{99}{10^2} \)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.......\dfrac{9.11}{10.10}\)
\(A=\dfrac{31}{30}.\dfrac{1}{2}\)
\(A=\dfrac{31}{60}\)
Đúng 0
Bình luận (0)