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\(\frac{3x}{4.7}+\frac{3x}{7.10}+\frac{3x}{10.13}+\frac{3x}{13.16}+...+\frac{3x}{19.22}=\frac{-5}{88}\)
\(\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{19.22}\right)x=\frac{-5}{88}\)
\(\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{19}-\frac{1}{22}\right)x=\frac{-5}{88}\)
\(\left[\frac{1}{4}+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{22}\right]x=\frac{-5}{88}\)
\(\left[\frac{1}{4}-\frac{1}{22}\right]x=\frac{-5}{88}\)
\(\frac{9}{44}x=\frac{-5}{88}\)
\(x=\frac{-5}{88}:\frac{9}{44}\)
\(x=\frac{-5}{18}\)
~ Hok tốt ~
#)Giải :
Đặt \(A=\frac{3x}{2.7}+\frac{3x}{7.10}+\frac{3x}{10.13}+\frac{3x}{13.16}+...+\frac{3x}{19.22}=-\frac{5}{88}\)
\(A=\frac{3x}{2}+\frac{3x}{7}-\frac{3x}{7}+\frac{3x}{10}-\frac{3x}{10}+\frac{3x}{13}-\frac{3x}{13}+\frac{3x}{16}-...-\frac{3x}{19}+\frac{3x}{22}=-\frac{5}{88}\)
\(A=\frac{3x}{2}+0+0+0+...+0+\frac{3x}{22}=-\frac{5}{88}\)
\(A=\frac{3x}{2}+\frac{3x}{22}=-\frac{5}{88}\)
\(3x:\left(2+22\right)=-\frac{5}{88}\)
\(3x:24=-\frac{5}{88}\)
\(3x=-\frac{5}{88}.24\)
\(3x=-\frac{7}{11}\)
\(x=-\frac{7}{11}:3\)
\(x=-\frac{7}{33}\)
#~Will~be~Pens~#
=>\(3A=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{91.94}+\frac{3}{94.97}\)
=>\(3A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{13}+...+\frac{1}{91}-\frac{1}{94}+\frac{1}{94}-\frac{1}{97}\)
=>\(3A=1-\frac{1}{97}\)
=>3A=\(\frac{96}{97}\)
=>A=\(\frac{32}{97}\)
\(S=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{2002\cdot2005}\)
\(3S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{2002\cdot2005}\)
\(3S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)
\(3S=\frac{1}{1}-\frac{1}{2005}\)
\(3S=\frac{2004}{2005}\)
\(S=\frac{2004}{2005}\div3=\frac{668}{2005}\)
Ta có:
\(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2002.2005}\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2002.2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2002}-\frac{1}{2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{2005}\right)=\frac{1}{3}.\frac{2004}{2005}=\frac{668}{2005}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(S=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
\(T=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(T=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+....+\frac{43-40}{40.43}+\frac{46-43}{43.46}\)
\(T=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(T=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)
1) Ta có : \(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy T = \(=\frac{99}{100}\)
2) Ta có : \(T=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}=\frac{45}{46}\)
Vậy T = \(\frac{45}{46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Đến đây ta suy được ra S<1
Ta có :
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy \(S< 1\)
Võ Thiện Tuấn viết tổng quát kết quả hay phép đề bài hả bạn ?
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7} +....+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}\)
\(=\frac{102}{103}\)
\(\frac{x}{1.4}+\frac{x}{4.7}+\frac{x}{7.10}+...+\frac{x}{36.39}=1\)
\(\frac{x}{3}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{36.39}\right)=1\)
\(\frac{x}{3}.[(\frac{1}{1}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+...+(\frac{1}{36}-\frac{1}{39})]=1\)
\(\frac{x}{3}.(\frac{1}{1}-\frac{1}{39})=1\)
\(\frac{x}{3}.\frac{38}{39}=1\)
\(\frac{x}{3}=1:\frac{38}{39}\)
\(\frac{x}{3}=\frac{39}{38}\)
\(\Rightarrow x=.....\)
Mình tính vội nên không tính kết quả đúng chưa, cậu kiểm tra lại nha, còn cách làm thế là chuẩn rồi! Học tốt!
1.
a) \(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\\ A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\\ A=1-\frac{1}{100}=\frac{99}{100}\)
b) Sửa đề: B = 1/2.5 + 1/5.8 + 1/8.11 + ...
\(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\\ B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\\ B=\frac{1}{6}-\frac{1}{294}\\ B=\frac{49}{294}-\frac{1}{294}=\frac{48}{294}=\frac{8}{49}\)
2.
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ \frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ 2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{2000}:2\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4000}\\ \frac{1}{2}-\frac{1999}{4000}=\frac{1}{n+1}\\ \frac{1}{n+1}=\frac{1}{4000}\\ \Rightarrow n+1=4000\\ \Rightarrow n=3999\)
Vậy n = 3999
Sửa đề nhé: \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{19.22}\)
\(S=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{19.22}\right)\)
\(S=\frac{1}{3}\left(\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{22-19}{19.22}\right)\)
\(S=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\right)\)
\(S=\frac{1}{3}\left(1-\frac{1}{22}\right)\)
\(S=\frac{1}{3}.\frac{21}{22}\)
\(S=\frac{7}{22}\)