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\(\Rightarrow A=\frac{14}{15}.\frac{20}{21}.\frac{41}{42}.....\frac{209}{210}\)
\(=\frac{4.7}{5.6}.\frac{5.8}{6.7}.\frac{6.9}{7.8}.....\frac{19.22}{20.21}\)
\(=\frac{22}{6}=\frac{11}{3}\)
=\(-\frac{6}{5}\).\(\frac{-7}{6}\).\(\frac{-8}{7}\).\(\frac{-9}{8}\).\(\frac{-10}{9}\).\(\frac{-11}{10}\)
=\(\frac{7}{5}\).\(\frac{9}{7}\).\(\frac{11}{9}\)
=\(\frac{11}{5}\)
\(=\frac{-6}{5}\times\frac{-7}{6}\times\frac{-8}{7}\times\frac{-9}{8}\times\frac{-10}{9}\times\frac{-11}{10}\)
\(=\frac{\left(-6\right).\left(-7\right).\left(-8\right).\left(-9\right).\left(-10\right).\left(-11\right)}{5.6.7.8.9.10}\)
\(=\frac{6\times7\times8\times9\times10\times11}{5\times6\times7\times8\times9\times10}\)
Triệt tiêu các thừa số bằng nhau ở tử và mẫu, ta có kết quả là \(\frac{11}{5}\)
\(\left(-2\right).\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2013}\right)\)
\(=\left(-2\right).\left(\frac{-3}{2}\right)\left(-\frac{4}{3}\right)......\left(\frac{-2014}{2013}\right)\)
\(=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2014\right)}{2.3.....2013}\)
\(=\frac{2.3.4....2014\left(\text{Vì có 2014 thừa số âm }\right)}{2.3....2013}\)
\(=\frac{\left(2.3.4....2013\right).2014}{2.3....2013}\)
\(=2014\)
\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{100}-1\right)\)
\(-A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{100}\right)\)
\(-A=\frac{9}{10}\cdot\frac{10}{11}\cdot\frac{11}{12}\cdot...\cdot\frac{99}{100}\)
\(-a=\frac{9}{100}\)
\(A=-\frac{9}{100}\)
Bài 1.
Ta có: \(\frac{a}{b}+\frac{-a}{b+1}=\frac{a}{b}-\frac{a}{b+1}=a\left(\frac{1}{b}-\frac{1}{b+1}\right)=a\left(\frac{b+1-b}{b\left(b+1\right)}\right)=a\left(\frac{1}{b\left(b+1\right)}\right)=\frac{a}{b\left(b+1\right)}\)
=> A là đáp án đúng
Bài 2. Ta có:
B = 4x - 4y + 5xy
B= 4x - 4y + 4xy + xy
B = 4(x - y + xy) + xy
B = 4.(5/12 - 1/3) - 1/3
B = 4.1/12 - 1/3
B = 1/3 - 1/3 = 0
\(\left(1+\frac{1}{1}\right).\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{5}\right).\left(1+\frac{1}{6}\right)\)
\(=2.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.\frac{7}{6}\)
\(=\frac{2.3.4.5.6.7}{2.3.4.5.6}=7\)