\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)

\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)

@seven

a: 1/1*2+1/2*3+...+1/99*100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23

=1/11-1/23

=12/253

\(B=\dfrac{40404}{70707}+\dfrac{244\times395-151}{244+395\times243}+\dfrac{1\times3\times5+2\times6\times10+4\times12\times20+7\times21\times35}{1\times5\times7+2\times10\times14+4\times20\times28+7\times35\times49}\\ =\dfrac{4}{7}+\dfrac{243\times395+395-151}{244+395\times243}+\dfrac{1\times3\times5\left(1+2+4+7\right)}{1\times5\times7\left(1+2+4+7\right)}\\ =\dfrac{4}{7}+\dfrac{243\times395+244}{244+395\times243}+\dfrac{3}{7}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+1\\ =1+1=2\)

a) \(\dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{3\times2\times5\times25\times7\times8}{25\times3\times8\times3\times4\times2\times7}=\dfrac{5}{3\times4}=\dfrac{5}{12}\)

b) \(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{8\times2\times3}=2\)

c) \(\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{4\times5\times3\times2}{3\times5\times2\times4\times2}=\dfrac{1}{2}\)

giúp mik với ạ, mình sẽ tick ạ! Thanks.

giúp mình với! Thanks

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004\)

1/ Ta có :

\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{49\times50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{50}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\times2\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)

\(\Rightarrow\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}\)

Hay \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...+\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)

~ Học tốt nha ~

\(M=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(M=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

\(M=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(M=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)

Giúp mik zới, mik đang cần gấp

bài 1 : rút gọn

bài 2 : tìm x

Đặt A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

=> A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=> A = 1 - \(\dfrac{1}{100}\) = \(\dfrac{99}{100}\)

=> 1 = \(\dfrac{100}{100}\)

=> A < 1

A =

=> A =

=> A = 1 - =

=> 1 =

=> A < 1

\(B=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)

\(=\left(\dfrac{1}{1.2}+\dfrac{4}{1.2}\right)+\left(\dfrac{1}{2.3}+\dfrac{12}{2.3}\right)+\left(\dfrac{1}{3.4}+\dfrac{24}{3.4}\right)+...+\left(\dfrac{1}{9.10}+\dfrac{180}{9.10}\right)\)

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)+\left(\dfrac{4}{1.2}+\dfrac{12}{2.3}+...+\dfrac{180}{9.10}\right)\)

\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(2+2+...+2\right)\)

\(=1-\dfrac{1}{10}+\left(2.9\right)\)

\(=1-\dfrac{1}{10}+18\)

\(=\dfrac{9}{10}+18\)

\(=18\dfrac{9}{10}\)