Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)
\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)
\(\left(x+1\right)=1:\dfrac{-251}{1006}\)
\(x+1=\dfrac{-1006}{251}\)
\(x=\dfrac{-1006}{251}-1\)
\(x=\dfrac{-1257}{251}\)
Nếu bạn tìm \(x\in Z\) hay \(x\in N\) thì \(x=\varnothing\) (không có x thoả mãn)
\(\Leftrightarrow2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1005}{1006}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-251}{1006}\)
=>x+1=-1006/251
hay x=-1257/251
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)
\(\Leftrightarrow\)\(2^x.15=480\)
\(\Rightarrow\)\(2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5.
\(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=17\)
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
\(\left(1\dfrac{3}{4}-\dfrac{4}{6}\right):\left(1\dfrac{1}{5}+2\dfrac{2}{5}+\dfrac{1}{5}\right)< x< 1\dfrac{1}{5}.1\dfrac{1}{4}+3\dfrac{2}{11}:2\dfrac{3}{121}\)
\(\Leftrightarrow\left(\dfrac{7}{4}-\dfrac{4}{6}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)< x< \dfrac{6}{5}.\dfrac{5}{4}+\dfrac{35}{11}:\dfrac{245}{121}\) \(\Leftrightarrow\left(\dfrac{21}{12}-\dfrac{8}{12}\right):\dfrac{19}{5}< x< \dfrac{3}{2}+\dfrac{35}{11}.\dfrac{121}{245}\) \(\Leftrightarrow\dfrac{13}{12}.\dfrac{5}{19}< x< \dfrac{3}{2}+\dfrac{2}{7}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{21}{14}+\dfrac{4}{14}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{25}{14}\) \(\Leftrightarrow x=1\)
a)
\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{x+3}=\dfrac{9}{38}\\\\ \dfrac{1}{x+3}=\dfrac{1}{4}-\dfrac{9}{38}\\ \dfrac{1}{x+3}=\dfrac{1}{76}\\ x+3=76\\ x=73.\)
b)
\(\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ 2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{1}{18}\\ x+1=18\\ x=17.\)
https://hoc24.vn/question/246430.html
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)
\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)
\(\left(x+1\right)=1:\dfrac{-251}{1006}\)
\(x+1=\dfrac{-1006}{251}\)
\(x=\dfrac{-1006}{251}-1\)
\(x=\dfrac{-1257}{251}\)
Vì \(x\in N\) nên \(x=\varnothing\) (không có giá trị nào của x thoả mãn)