1 + 2/2x3 + 2/3x4 + ....+ 2/198x199 + 1/199x100 = ?
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\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)
\(P=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(P=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
Bài 1:
Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)
\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)
\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)
\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(A=\frac{2x19}{20}=\frac{19}{10}\)
Bài 2:
Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)
Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)
\(Bx100=\frac{9}{10}x100=90\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)
=> \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)
=2(1/1.2+1/2.3 +1/3.4 +...+1/199.200)
= 2.(1-1/2 +1/2-1/3 + ...+1/199-1/200)
=2.199/200 = 199/100