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đề có sai không đó bạn làm gì tích 2 số tự nhiên liên mà = 900 chớ
1/1x2+1/2x3+...+1/49x50
=1-1/2+1/2-1/3+.....+1/49-1/50
=1-1/50(1)
Ta co 1(2)
So sanh (1) voi (2) ta thay 1-1/50<1
=>1/1x2+...+1/49x50<1
(Phuong phap khu)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)
Vậy \(\frac{49}{50}<1\)
a) \(\frac{3}{40}+\frac{5}{3}+\frac{7}{60}=\frac{9}{120}+\frac{200}{120}+\frac{14}{120}=\frac{223}{120}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{19.20}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\Rightarrowđpcm\)
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
Mà : \(\frac{99}{100}< 1\)
Vậy : S < 1
Tổng trên = 1-2^2/2^2 . 1-3^2/3^2 . ..... . 1-100^2/100^2
= -(2^2-1/2^2 . 3^2-1/3^2 . ...... . 100^2-1/100^2 )
= -(1.3/2^2 . 2.4/3^2 . ..... . 99.101/100)
= -(1.2.3. .... .99 . 3.4.5. ... .101 / 2.3.4 . ... . 100 . 2.3.4 . ..... . 100)
= -(1.2.3. ... . 99/2.3.4. .... .100) . (3.4.5. .... .101/2.3.4 . .... . 100)
= -1/100 . 101/2 = -101/200
Tk mk nha
H = \(\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)
Đặt G = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Đặt K = \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)
=>2K = \(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{99.100.101}\right)\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)
= \(\frac{1}{1.2}-\frac{1}{100.101}\)
= \(\frac{1}{2}-\frac{1}{10100}\)
= \(\frac{5049}{10100}\)
=> K =\(\frac{5049}{10100}:2=\frac{5049}{10100}.\frac{1}{2}=\frac{5049}{20200}\)
Thay G,K vào H ta có :
H = \(\frac{99}{100}-\frac{5049}{20200}\)
Tự tính :)
\(H=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.34}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{99}{100}-\frac{1}{2}.\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
a) \(\frac{2^{12}x3^5-4^6.9^2}{\left(2^2x3\right)^6+8^4x3^5}=\frac{2^{12}x3^5+\left(2^2\right)^6x\left(3^2\right)^2}{2^{12}x3^6+\left(2^3\right)^4x3^5}\)
\(=\frac{2^{12}x3^5-2^{12}x3^4}{2^{12}x3^6+2^{12}x3^5}=\frac{2^{12}x3^4x\left(3-1\right)}{2^{12}x3^5x\left(3+1\right)}\)
\(=\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)
b) \(\frac{1}{9x10}-\frac{1}{8x9}-\frac{1}{7x8}-\frac{1}{6x7}-\frac{1}{5x6}-\frac{1}{4x5}-\frac{1}{3x4}-\frac{1}{2x3}-\frac{1}{1x2}\)
\(=-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(1-\frac{1}{10}\right)\)
\(=\frac{-9}{10}\)
sorry bn nha! mk ko bk lm phần c
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=1-\frac{1}{2011}\)
\(S=\frac{2010}{2011}\)
=1+2x2+3x3+4x...x2010+2011 =1x2+2x3+3x4+...+2010x2011 =1x2011 =2011