Tìm \(x\):
\(\dfrac{x-2}{3.4}+\dfrac{x-2}{4.5}+\dfrac{x-2}{5.6}+...+\dfrac{x-2}{8.9}=\dfrac{16}{9}\)
K
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NT
1
Những câu hỏi liên quan
6 tháng 4 2021
3.(1/4.5+1/5.6+...+1/9.10).x=9/2
3.(1/4-1/5+1/5-1/6+...+1/9-1/10).x=9/2
3.(1/4-1/10).x=9/2
3.3/20.x=9/2
9/20.x=9/2
x=9/2:9/20
x=10
28 tháng 4 2021
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
MH
6 tháng 5 2022
\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)
\(\left(x+1\right)^2=324=18^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=18\\x+1=-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-19\end{matrix}\right.\)
6 tháng 5 2022
Ta có \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{x+1}{324}+\dfrac{1}{x+1}\)
\(\Rightarrow\)\(\dfrac{1}{x+1}-\dfrac{x+1}{324}=0\)
\(\Rightarrow\)\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)
\(\Rightarrow\)(x+1).(x+1)=324
\(\Rightarrow\)(x+1)2=324
\(\Rightarrow\)(x+1)2 = 182 = (-18)2
TH1: (x+1)2 = 182
\(\Rightarrow\)x+1 = 18
\(\Rightarrow\)x = 17
TH2: (x+1)2 = (-18)2
\(\Rightarrow\)x+1 = -18
\(\Rightarrow\)x = -19
Vậy x\(\in\)\(\left\{17;-19\right\}\)
T
5 tháng 3 2017
đặt \(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)
ta có:
\(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)
\(\Leftrightarrow M=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}\) \(\Leftrightarrow M=\dfrac{3}{3.4}+\dfrac{4}{3.4}-\dfrac{4}{4.5}-\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{6}{5.6}-\dfrac{6}{6.7}-\dfrac{7}{6.7}+\dfrac{7}{7.8}+\dfrac{8}{7.8}-\dfrac{8}{8.9}-\dfrac{9}{8.9}+\dfrac{9}{9.10}+\dfrac{10}{9.10}\) \(\Rightarrow M=\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{8}+\dfrac{1}{10}+\dfrac{1}{9}\) \(\Rightarrow M=\dfrac{1}{3}+\dfrac{1}{10}\)
\(\Rightarrow M=\dfrac{10}{30}+\dfrac{3}{30}\)
\(\Rightarrow M=\dfrac{13}{30}\)
vậy M = \(\dfrac{13}{30}\)
vậy \(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{13}{30}\)
5 tháng 3 2017
\(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{7}{30}\)
ND
Nguyễn Đức Trí
VIP
21 tháng 7 2023
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 8 2023
B = \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\) - \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\)
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\))
B = \(\dfrac{1}{12}\) - \(\dfrac{3}{20}\)
B = - \(\dfrac{1}{15}\)
5 tháng 3 2018
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
16 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)
`=`\(\dfrac{2}{9}\)
Vậy, \(A=\dfrac{2}{9}\)
`b)`
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
Vậy, \(B=\dfrac{4}{25}\)
`c)`
\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)
`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy, \(C=\dfrac{99}{100}\)
27 tháng 4 2018
a)
\(\dfrac{1}{2\cdot3}x+\dfrac{1}{3\cdot4}x+...+\dfrac{1}{49\cdot50}x=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ x\cdot\dfrac{12}{25}=1\\ x=1:\dfrac{12}{25}=1\cdot\dfrac{25}{12}=\dfrac{25}{12}\)
\(\dfrac{x-2}{3.4}+\dfrac{x-2}{4.5}+...+\dfrac{x-2}{8.9}=\dfrac{16}{9}\\ \\ \\ \\ \Rightarrow\left(x-2\right).\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{8.9}\right)=\dfrac{16}{9}\\ \\ \\ \\ \Rightarrow\left(x-2\right).\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \\ \\ \\ \Rightarrow\left(x-2\right).\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \\ \\ \\ \Rightarrow\left(x-2\right).\dfrac{2}{9}=\dfrac{16}{9}\\ \\ \\ \\ \Rightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=8\\ \\ \\ \\ \Rightarrow x=8+2=10\)
Vậy \(x=10\)