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(1-1/3)x(1-1/5)x(1-1/7)x(1-1/9)x(1-1/2)x(1-1/4)x(1-1/6)x(1-1/8)x(1-1/10)
=2/3x4/5x6/7x8/9x1/2x3/4x5/6x7/8x9/10
=2x4x6x8x1x3x5x7x9 /3x5x7x9x2x4x6x8x10
=1/10
c) =-5/7x3/9+4/9x3/-7
=-5/7x3/9+3/9x4/-7
=3/9x(-5/7+4/-7)
=3/9x-9/7
=-3/7
d) =58/7 - (31/9+30/7)
=58/7-31/9-30/7
=(58/7-30/7)-(31/9)
=4-31/9
=5/9
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(a=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
\(a=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(a=1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}\)
Vậy \(a=\frac{99}{100}\)