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(1+1/2)x(1+1/3)x(1+1/4)x....x(1+1/98)x(1+1/99)
=3/2x4/3x5/4x..........x99/98x100/99
=(3x4x5x........x99x100)/2x3x4x........x98x99
Giản ước ta được:
=100/2
=50
(1-1/2)x(1-1/3)x(1-1/4)x.......x(1-1/2007)
giải chi tiêt giùm mình nha!
mình chỉ biết đáp án là 1/2007
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x............x\left(1-\frac{1}{2017}\right)\))
= \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x..............x\frac{2006}{2007}\)
= \(\frac{1}{2007}\)
Đó là kết quả sau khi trực tiêu
Bài này dễ mà
Trong violympic
(1+1/2)(1+1/3)...(1+1/99)
=(3/2)(4/3)(5/4)...(100/99)
=100/2=50
\(B=\left(1+\dfrac{1}{100}\right)\times\left(1+\dfrac{1}{99}\right)\times....\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{2}\right)\)
\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)
\(B=\dfrac{101\times100\times....\times4\times3}{100\times99\times....\times3\times2}\)
\(B=\dfrac{101}{2}\)
\(\Rightarrow B=\left(\dfrac{100}{100}+\dfrac{1}{100}\right)\times\left(\dfrac{99}{99}+\dfrac{1}{99}\right)\times...\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)
\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)
\(B=\dfrac{101}{2}\)( triệt tiêu các mẫu, tử giống nhau)
= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10
= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10
=1-1/10=9/10
đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10
\(A=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+...+\frac{1}{9}\cdot\frac{1}{10}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100
\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2\times\frac{99}{100}\)
\(=\frac{99}{50}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
học tốt nha
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{98}{99}\times\dfrac{99}{100}\)
\(=\dfrac{1\times2\times3\times...\times98\times99}{2\times3\times4\times...\times99\times100}=\dfrac{1}{100}\)
(1 - 1/2) × (1 - 1/3) × (1 - 1/4) × ... × (1 - 1/99) × (1 - 1/100)
= 1/2 × 2/3 × 3/4 × ... × 98/99 × 99/100
= 1/100