\(x\dfrac{x}{15}=\dfrac{112}{5}\)Tìm x.
Lời giải thật chi tiết nha.
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 7 2023
\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)
\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)
\(x\) + 1 = 16
\(x\) = 16 - 1
\(x\) = 15
XO
16 tháng 7 2023
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)
\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 7 2023
Em giải như XYZ olm em nhé
Sau đó em thêm vào lập luận sau:
\(x\) = \(\dfrac{11}{18}\)
Vì \(\in\) N*
Vậy \(x\in\) \(\varnothing\)
24 tháng 3 2022
dấu . là nhân hay là phần ngăn cách ở hàng phần nghìn thế
BD
6 tháng 7 2023
\(\dfrac{1}{3}+\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}\)
\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}\cdot\dfrac{6}{5}\)
\(x-\dfrac{11}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{11}{5}\)
\(x=\dfrac{27}{10}\)
6 tháng 7 2023
\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}:\dfrac{5}{6}\)
\(x-\dfrac{11}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{11}{5}=\dfrac{27}{10}\)
KH
5 tháng 7 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\x-2>0\\x+2>0\\x\ge0\end{matrix}\right.\) và \(4-x\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x>2\\x>-2\\x\ge0\end{matrix}\right.\) và \(x\ne4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)
K
24 tháng 3 2022
\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)
23 tháng 2 2022
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)
DT
23 tháng 2 2022
\(\Leftrightarrow\dfrac{1}{4}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{\left(x-1\right)x}\right)=\dfrac{1}{8}\) ( đk x khác 0 , x khác 1)
\(\Leftrightarrow\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{x-1}-\dfrac{1}{x}\right)=\dfrac{1}{8}\)
\(\Leftrightarrow1-\dfrac{1}{x}=\dfrac{1}{2}\)
=> x =2 ( tm)
Có:
\(x\dfrac{x}{15}=\dfrac{112}{5}\)
Hay \(\dfrac{x.15+x}{15}=\dfrac{112}{5}\)
\(\Rightarrow x.15+x=\dfrac{112.15}{5}=336\)
\(x.\left(15+1\right)=336\)
\(x.16=336\)
\(\Rightarrow x=\dfrac{336}{16}=21\)
Chúc bạn học tốt!
\(x\dfrac{x}{15}=\dfrac{112}{5}\)
\(\Rightarrow\dfrac{15x+x}{15}=\dfrac{112}{5}\)
\(\Rightarrow\dfrac{15x+x}{15}=\dfrac{336}{15}\)
\(\Rightarrow15x+x=336\)
\(16x=336\)
\(x=336:16\)
\(x=21\)
Vậy \(x=21\)