\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{120}{121}=\frac{3.8.15...120}{4.9.16...121}\)

    \(=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)...\left(11.11\right)}\)

    \(=\frac{\left(1.2.3...10\right).\left(3.4.5...12\right)}{\left(2.3.4...11\right).\left(2.3.4...11\right)}=\frac{1.12}{11.2}=\frac{6}{11}\)

ta có :

A=\(\left(-\frac{3}{4}\right)\left(-\frac{8}{9}\right)\left(-\frac{15}{16}\right)...\left(-\frac{120}{121}\right)\)(có 10 số hạng)

  = \(\frac{3\cdot8\cdot15\cdot...\cdot120}{4\cdot9\cdot16\cdot...\cdot121}=\frac{\left(1.3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(10\cdot12\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot11^2}=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot..\cdot11\right)\left(2\cdot3\cdot4\cdot..\cdot11\right)}\)

=\(\frac{12}{11\cdot2}=\frac{12}{22}\)

A =\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)...\left(\frac{1}{121}-1\right)\)

A \(=\frac{-3}{4}.\frac{\left(-8\right)}{9}.\frac{\left(-15\right)}{16}.\frac{\left(-24\right)}{25}...\frac{\left(-120\right)}{121}\)

A \(=\frac{-3}{2^2}.\frac{2.\left(-4\right)}{3^2}.\frac{\left(-3\right).5}{4^2}.\frac{\left(-4\right).6}{5^2}...\frac{10.\left(-12\right)}{11^2}\)

A \(=\frac{2.3^2.4^2.5^2.6^2...10^2.\left(-12\right)}{2^2.3^2.4^2.5^2...11^2}=\frac{-12}{2}=-6\)

S=\(^{2^{2010}-2^{2009}-2^{2008}-...-2-1}\)

Gộp nhóm 4 => A = -4 * 500+2001+2002-2003=0

B = 

a) tạm bỏ số 1 ra => có 2012 số hạng=> có 1006 cặp =(-1)

=> A=1+-(-1).1006=-1005

A=-3.(1-(2/3)²)(1-(2/5)²)...(1-(2/11)²)=-3.(1-2/3)(1+2/3)(1-2/5)(1+2/5)...(1-2/11)(1+2/11)=-3.\(\frac{1}{3}\).\(\frac{5}{3}\).\(\frac{3}{5}\).\(\frac{7}{5}\)...\(\frac{9}{11}.\frac{13}{11}\)

=  -\(\frac{13}{11}\)