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\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left[x+1\right]}=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2017}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{\frac{2017}{2019}}{2}=\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}=\frac{1}{2019}\)
=> x + 1 = 2019 <=> x = 2018
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Leftrightarrow x=308-3\)
\(\Leftrightarrow x=305\)
Vậy \(x=305\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{2017}{2019}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2019}\)
\(\Leftrightarrow x+1=2019\)
\(\Leftrightarrow x=2019-1\)
\(\Leftrightarrow x=2018\)
Vậy x = 2018
chị ơi bài này em học từ lớp 6 rồi( ngay đầu vào luôn ) nhung dốt nát lè
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x-1\right)}=\)\(\frac{2017}{2019}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x-1\right)}=\frac{2017}{2019}\)
\(2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right]=\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\)\(\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{2019}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2019}\)
x + 1 =2019
x = 2019-1 =2018
Vậy x = 2018
\(2\left(\frac{1}{3}.\frac{1}{2}+\frac{1}{6}.\frac{1}{2}+\frac{1}{10}.\frac{1}{2}+....+\frac{2}{x\left(x+1\right)}.\frac{1}{2}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)\)\(=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2[\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+....+\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+1}]=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}+0+0+....+0-\frac{1}{x-1}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=>\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{4038}\)
=>\(\frac{1}{x+1}=\frac{1}{2019}\)
=> x+1=2019
=>x=2018
Bạn tham khảo câu trả lời tương tự ở đây nhé:
Câu hỏi của Nguyễn Hải - Toán lớp 7 - Học toán với OnlineMath
\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
\(\frac{2}{6}\)+\(\frac{2}{12}\)+\(\frac{2}{20}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{n.\left(n+1\right)}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{3}\)+\(\frac{1}{3}\)_\(\frac{1}{4}\)+\(\frac{1}{4}\)_\(\frac{1}{5}\)+...+\(\frac{1}{n}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
\(\frac{1}{2}\)_\(\frac{1}{n+1}\)=\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2}\)_\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2019}\)
\(\Rightarrow\)n+1=2019
\(\Rightarrow\)n=2018\(\in\)Z
Vậy n=2018
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)
\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)
\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)
\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)
\(\Rightarrow-480\le x\le480\)
\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)
\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)
\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)
\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)
\(\Leftrightarrow x-3=\frac{90}{7}\)
\(\Rightarrow x=\frac{111}{7}\)
\(\frac{\frac{25}{108}.\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)=\(\frac{\frac{5755}{108}+\frac{187}{4}}{\frac{-973}{410}}\)=\(\frac{\frac{8531}{84}}{\frac{-973}{410}}\)=-241,0180
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(=\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right)\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\left(-\frac{41}{21}\right)}\)
\(=\frac{\frac{25}{108}\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}\cdot\left(-\frac{21}{41}\right)}=\frac{\frac{2701}{27}}{-\frac{973}{410}}\)
Tính nốt vì số dữ quá , lần sau để số ít thôi
\(1.\left(x+3\right)^3=\frac{1}{-27}\)
\(\left(x+3\right)^3=\left(\frac{1}{-3}\right)^3\)
\(\Rightarrow x+3=\frac{1}{-3}\)
\(\Rightarrow x=\frac{-1}{3}-3\)
\(x=\frac{-10}{3}\)