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\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{120}{121}=\frac{3.8.15...120}{4.9.16...121}\)
\(=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)...\left(11.11\right)}\)
\(=\frac{\left(1.2.3...10\right).\left(3.4.5...12\right)}{\left(2.3.4...11\right).\left(2.3.4...11\right)}=\frac{1.12}{11.2}=\frac{6}{11}\)
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, Đúng rồi đó
b, \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)+....+\left(1+\frac{1}{39.41}\right)\)
= \(\frac{4}{1.3}.\frac{9}{2.4}.....\frac{1600}{39.41}\)
= \(\frac{2.2.3.3....40.40}{1.3.2.4....39.41}\)
= \(\frac{\left(2.3....40\right)\left(2.3....40\right)}{\left(1.2....39\right)\left(3.4....41\right)}\)
= \(\frac{40.2}{41}\)
= \(\frac{80}{41}\)
\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{8^2}\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{8^2-1}{8^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{7.9}{8.8}\)
\(=\frac{\left(1.2.....7\right).\left(3.4.....9\right)}{\left(2.3.....8\right).\left(2.3.....8\right)}\)
\(=\frac{1.9}{8.2}=\frac{9}{16}\)