Question

[Trần Ngọc Lan Anh]

Tính giá trị biểu thức: 

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

Answer 1

[Không Tên]

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

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

\(=\frac{3}{2^2}.\frac{8}{3^2}......\frac{9603}{98^2}.\frac{9800}{99^2}\)

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{97.99}{98^2}.\frac{98.100}{99^2}\)

\(=\frac{1.2.4...97.98}{2.3....98.99}.\frac{3.4...99.100}{2.3....98.99}\)

\(=\frac{1}{99}.\frac{100}{2}\)

\(=\frac{50}{99}\)

Answer 2

[Trần Ngọc Lan Anh]

bn viết sai 1 chỗ nhưng ko s ^^ tks nhoa