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Bài 58:
a, \(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)
b, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)
Vậy...
\(B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}\)
\(=\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}\)
\(=\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+7\right)+6\left(x+7\right)}\)
\(=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}\)
\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}\)
\(=\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}\)
\(=\dfrac{3}{x^2+11x+28}\)
Vậy...
58,
\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)b,
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)\(=1-\dfrac{1}{n\left(n+1\right)}=\dfrac{n^2+n-1}{n\left(n+1\right)}\)
\(B=\dfrac{1}{\left(x^2+9x+20\right)}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}\)\(=\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\dfrac{1}{\left(x^2+6x\right)+\left(7x+42\right)}\)\(=\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}\)\(=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x-7}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+7}\)
Đặt: \(\left\{{}\begin{matrix}l_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2005.2006.2007}\\l_2=1.2+2.3+3.4+...+2006.2007\end{matrix}\right.\Leftrightarrow l_1.x=l_2\)
Ta có:
\(l_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2005.2006.2007}\)
\(l_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2005.2006}-\dfrac{1}{2006.2007}\right)\)
\(l_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2006.2007}\right)\)
\(l_2=1.2+2.3+3.4+...+2006.2007\)
\(3l_2=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2006.2007.\left(2008-2005\right)\)
\(3l_2=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2006.2007.2008-2005.2006.2007\)
\(3l_2=2006.2007.2008\Leftrightarrow l_2=\dfrac{2006.2007.2008}{3}\)
Hay: \(\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2006.2007}\right)\right].x=\dfrac{2006.2007.2008}{3}\)
Tới đây thì bấm máy tính là ra :V
Nhã Doanh, ngonhuminh, nguyen thi vang, Hoàng Anh Thư, Mashiro Shiina, Phạm Nguyễn Tất Đạt, F.C, Trần Thị Hồng Ngát, Mến Vũ, kuroba kaito, @Phùng Khánh Linh, Nguyễn Huy Tú, Lightning Farron, Hung nguyen, ...
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .
\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)
\(A=1-\dfrac{1}{n^2+2n+1}\)
\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)
Ta biến đổi \(A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{2016-2015}{2016.2015}+\dfrac{2018-2017}{2017.2018}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1009}\right)\)
\(A=\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2017}+\dfrac{1}{2018}\)
Lại có \(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\)
\(B=\dfrac{1}{3028}.\left(\dfrac{3028}{1010.2018}+\dfrac{3028}{1011.2017}+...+\dfrac{3028}{2018.1010}\right)\)
\(B=\dfrac{1}{3028}\left(\dfrac{1}{1010}+\dfrac{1}{2018}+\dfrac{1}{1011}+\dfrac{1}{2017}+...+\dfrac{1}{2018}+\dfrac{1}{1010}\right)\)
\(B=\dfrac{1}{3028}.2\left(\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2018}\right)\)
\(B=\dfrac{1}{3028}.2A\) \(\Rightarrow\dfrac{A}{B}=1514\inℤ\). Ta có đpcm
\(L_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2015.2016.2017}\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{4}-\dfrac{1}{2.2016.2017}\)
\(L_2=1.2+2.3+...+2006.2007\)
\(3L_2=1.2.3+2.3.\left(4-1\right)+...+2006.2007.\left(2008-2005\right)\)
\(3L_2=1.2.3+2.3.4-1.2.3+...+2006.2007.2008-2005.2006.2007\)\(3L_2=2006.2007.2008\)
\(L_2=\dfrac{2006.2007.2008}{3}\)
\(pt\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{2.2016.2017}\right).x=\dfrac{2006.2007.2008}{3}\)
Dễ dàng tìm được x nhé
Câu 1:
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
=>n+1=3000
hay n=2999
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..................+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+..................+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.......+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{99}-\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+.......+\dfrac{1}{100}\)
\(\dfrac{x}{1007}-\dfrac{1}{1.2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-...-\dfrac{1}{13.14}=\dfrac{15}{14}\)
⇔ \(\dfrac{x}{1007}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\right)=\dfrac{15}{14}\)
⇔ \(\dfrac{x}{1007}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{13}-\dfrac{1}{14}\right)=\dfrac{15}{14}\)
⇔ \(\dfrac{x}{1007}-\left(1-\dfrac{1}{14}\right)=\dfrac{15}{14}\)
⇔ \(\dfrac{x}{1007}-\dfrac{13}{14}=\dfrac{15}{14}\)
⇔ \(\dfrac{x}{1007}=\dfrac{15}{14}+\dfrac{13}{14}\)
⇔ \(\dfrac{x}{1007}=\dfrac{28}{14}\)
⇔ \(\dfrac{x}{1007}=2\)
⇔ \(x=2.1007\)
⇔ \(x=2014\)
Vậy \(x=2014\)
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