Question

1)Tính

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

b) B= 2000( 2001⁹ + 2001⁸ +... + 2001² + 2001)¹ +2

Answer 1

a)Ta có:

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

\(=\left(\frac{1}{2.2}-1\right)\left(\frac{1}{3.3}-1\right)\left(\frac{1}{4.4}-1\right)....\left(\frac{1}{98.98}-1\right)\left(\frac{1}{99.99}-1\right)\)

\(=\left(-\frac{3}{2.2}\right).\left(-\frac{8}{3.3}\right).\left(-\frac{15}{4.4}\right)...\left(-\frac{9603}{98.98}\right).\left(-\frac{9800}{99.99}\right)\)

\(=\left[\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)\right].\frac{3}{2.2}.\frac{8}{3.3}.\frac{15}{4.4}...\frac{9603}{98.98}.\frac{9800}{99.99}\)

   |------------------------98 số -1--------------------|

\(=\left(-1\right)^{98}.\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)

\(=\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)

\(=\frac{1.3.2.4.3.5...95.97.98.100}{2.2.3.3.4.4...98.98.99.99}\)

Ta sẽ rút gọn các thừa số chung ở tử và mẫu

\(=\frac{1.100}{2.99.99}\)

\(=\frac{50}{9801}\)

Vậy \(A=\frac{50}{9801}\)

Answer 2

cho mik hỏi b­ước 3  chỗ \(\frac{3}{2.2}\)sai o duoi lai la\(\frac{3}{2.3}\)vay