Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(...=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}....\dfrac{97}{98}.\dfrac{98}{99}\)
\(=\dfrac{1}{99}\)
Lời giải:\(A=\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-4\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)
\(A=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
\(\Rightarrow A=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right)}{2.3.4....99.100}\)
\(\Rightarrow A=\dfrac{1}{100}\)
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,
\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
Ta có: \(B=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Vậy T = 50
\(T=\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{98}+1\right)\cdot\left(\dfrac{1}{99}+1\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}+\dfrac{3}{3}\right)\cdot\left(\dfrac{1}{4}+\dfrac{4}{4}\right)\cdot...\cdot\left(\dfrac{1}{98}+\dfrac{98}{98}\right)\cdot\left(\dfrac{1}{99}+\dfrac{99}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{2\cdot3\cdot4\cdot...\cdot98\cdot99}\)
\(=\dfrac{100}{2}=50\).
1/ \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).........\left(1-\dfrac{1}{100}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}...............\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
2/ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+.........+\dfrac{1}{99.100}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+........+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{5}-\dfrac{1}{100}\)
\(=\dfrac{19}{100}\)
1. \(\left(1-\dfrac{1}{2}\right)\) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\) \(...\left(1-\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
2. \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{5}-\dfrac{1}{100}\)
\(=\dfrac{20}{100}\) \(-\dfrac{1}{100}\)
\(=\dfrac{19}{100}\)
\(\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{2^2}\right)+...+\left(1+\dfrac{1}{2^{50}}\right)\)
= \(\left(1+1+1+...+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)(50 số 1 )
= \(50+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)
A =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\)
⇒ 2A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\)
⇒ 2A - A =\(1-\dfrac{1}{2^{50}}\)
=50+1-\(\dfrac{1}{2^{50}}\)=51-\(\dfrac{1}{2^{50}}>3\)
\(B=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}=\dfrac{3.4.5...100}{2.3.4...99}=\dfrac{\left(3.4.5...99\right)100}{2\left(3.4.5...99\right)}=\dfrac{100}{2}=50\)
Vậy B = 50
\(B=\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{4}\right).....\left(1+\dfrac{1}{99}\right)\)
\(B=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{100}{99}\)
\(B=\dfrac{3.4.5....99.100}{2.3.4....98.99}\)
\(B=\dfrac{100}{2}\)
\(B=50\)
\(P=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{99}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{98}{99}\\ =\dfrac{1\cdot2\cdot3\cdot...\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{1}{99}\\ Vậy....\)