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\(T=\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).......\left(\dfrac{1}{98}+1\right).\left(\dfrac{1}{99}+1\right) \) \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}....\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
\(T=\left|\dfrac{1}{2}+1\right|\left|\dfrac{1}{3}+1\right|\left|\dfrac{1}{4}+1\right|.....\left|\dfrac{1}{98}+1\right|\left|\dfrac{1}{99}+1\right|\)
\(T=\left|\dfrac{3}{2}\right|.\left|\dfrac{4}{3}\right|.\left|\dfrac{5}{4}\right|......\left|\dfrac{99}{98}\right|.\left|\dfrac{100}{99}\right|\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5.....99.100}{2.3.4.....98.99}=\dfrac{100}{2}=50\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3....2014}{2.3.4....2015}\)
\(D=\dfrac{1}{2015}\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3...2014}{2.3.4...2015}\)
\(D=\dfrac{1}{2015}\)
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).........................\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\left(\dfrac{1}{4}-\dfrac{4}{4}\right)................\left(\dfrac{1}{99}-\dfrac{99}{99}\right)\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...................\left(\dfrac{-98}{99}\right)\left(\dfrac{-99}{100}\right)\)
\(A=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right).........................\left(-98\right)\left(-99\right)}{2.3.4....................98.99.100}\)
\(A=\dfrac{-1}{100}\)
Ta có
A = \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)(99 thừa số)
A = \(\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
A = \(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right).\left(-100\right)}{2.3.4....98.99.100}\)
A = \(\dfrac{\left(-1\right).\left(-1\right).\left(-1\right)....\left(-1\right)}{1.1.1...1.1.1}\) (100 số -1, 99 số 1)
A = \(\dfrac{-1}{1.1.1.1...1.1.1}\)
A = \(\dfrac{-1}{1}\)
A = -1
Vậy A = -1
\(A=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}=\dfrac{100}{2}=50\)
Vậy A = 50
Ta có:
A=\(^{\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).....\left(\dfrac{1}{99}\right)+1}\)
A= \(\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{1}{99}+1\)
A=\(\dfrac{1}{2}+1\)
A=\(\dfrac{3}{2}\)
\(A=\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\cdot...\left(1+\dfrac{1}{2499}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot...\cdot\dfrac{2500}{2499}\)
\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{50\cdot50}{49\cdot51}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot50}{1\cdot2\cdot3\cdot...\cdot49}\cdot\dfrac{2\cdot3\cdot...\cdot50}{3\cdot4\cdot...\cdot51}\)
\(=\dfrac{50}{1}\cdot\dfrac{2}{51}=\dfrac{100}{51}\)
a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)
\(=\dfrac{3.4.5...100}{2.3.4...99}\)
\(=\dfrac{100}{2}=50\)
a,
\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{100}{2}=50\)
b,
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-99}{100}\\ =\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)\cdot...\cdot\left(-99\right)}{2\cdot3\cdot4\cdot...\cdot100}\\ =\dfrac{\left(-1\right)\left(-1\right)\left(-1\right)\cdot...\left(-1\right)}{100}\left(\text{có }99\text{ số }-1\right)\\ =\dfrac{\left(-1\right)^{99}}{100}\\ =\dfrac{-1}{100}\)
c,
\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)
\(...=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}....\dfrac{97}{98}.\dfrac{98}{99}\)
\(=\dfrac{1}{99}\)
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Vậy T = 50
\(T=\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{98}+1\right)\cdot\left(\dfrac{1}{99}+1\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}+\dfrac{3}{3}\right)\cdot\left(\dfrac{1}{4}+\dfrac{4}{4}\right)\cdot...\cdot\left(\dfrac{1}{98}+\dfrac{98}{98}\right)\cdot\left(\dfrac{1}{99}+\dfrac{99}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{2\cdot3\cdot4\cdot...\cdot98\cdot99}\)
\(=\dfrac{100}{2}=50\).