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

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

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

\(=\left[\left(-1\right)\left(-1\right)\left(-1\right)\left(-1\right)....\left(-1\right)\left(10\right)\text{thừa số -1 }\right].\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{10.12}{11.11}\)

\(=\frac{1.12}{2.11}=\frac{6}{11}\)

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

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

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

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

=\(-\frac{11}{20}\)

\(S=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right).\left(\frac{1}{100}-1\right)\)

\(S=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}........\frac{-80}{81}.\frac{-99}{100}\)

\(-S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{80}{81}.\frac{99}{100}\)

\(-S=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}........\frac{8.10}{9.9}.\frac{9.11}{10.10}\)

\(-S=\frac{1.3.2.4.3.5........8.10.9.11}{2.2.3.3.4.4.......9.9.10.10}\)

\(-S=\frac{\left(1.2.3......8.9\right).\left(3.4.5.......10.11\right)}{\left(2.3.4.......9.10\right).\left(2.3.4........9.10\right)}\)\(=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}=>S=\frac{-11}{20}\)

sao câu này thầy ko chọn ạ

Ngân Hoàng Xuân: bị 2 lần rồi, ức chế v~~~