Question

[Trinh quang huy]

Tính A

A =\((\frac{1}{2^2}-1)\cdot(\frac{1}{3^2}-1)\cdot(\frac{1}{4^2}-1)...(\frac{1}{98^2}-1)\cdot(\frac{1}{99^2})\)

Answer 1

[Đỗ Thị Dung]

đề thiếu bn ơi

phải là \(\frac{1}{99^2}-1\)

A=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{98^2}-1\right)\).\(\left(\frac{1}{99^2}-1\right)\)

do tích A có: (99-2)+1=98 thừa số nguyên âm nên tích A dương

A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{97.99}{98^2}.\frac{98.100}{99^2}\)=\(\frac{1.3.2.4.3.5...97.99.98.100}{2^2.3^2.4^2...98^2.99^2}\)

=\(\frac{1.2.3.4...98}{2.3.4...98.99}.\frac{3.4.5...99.100}{2.3.4...98.99}\)=\(\frac{1}{99}.\frac{100}{2}\)=\(\frac{50}{99}\)

vậy A=\(\frac{50}{99}\)

#HỌC TỐT#