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T
1
TM
18 tháng 5 2016
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{18}\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{18}:\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}\)
<=>x+3=30
<=>x=27
Vậy x=27
LN
1
C
27 tháng 2 2023
\(\dfrac{11}{8}:x-\dfrac{2}{5}+-\dfrac{1}{6}=-\dfrac{1}{5}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}-\left(-\dfrac{1}{6}\right)\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}+\dfrac{1}{6}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{6}{30}+\dfrac{5}{30}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{30}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{2}{5}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{12}{30}\\ =>\dfrac{11}{8}:x=\dfrac{11}{30}\\ =>x=\dfrac{11}{8}:\dfrac{11}{30}\\ =>x=\dfrac{11}{8}.\dfrac{30}{11}\\ =>x=\dfrac{30}{8}\\ =>x=\dfrac{15}{4}\\ \dfrac{4}{7}x-\dfrac{1}{3}x+\left(-\dfrac{16}{21}\right)=-\dfrac{2}{3}\\ =>\left(\dfrac{4}{7}-\dfrac{1}{3}\right)x=-\dfrac{2}{3}-\left(-\dfrac{16}{21}\right)\\ =>\left(\dfrac{12}{21}-\dfrac{7}{21}\right)x=-\dfrac{2}{3}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=-\dfrac{14}{21}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=\dfrac{2}{21}\\ =>x=\dfrac{2}{21}:\dfrac{5}{21}\)
\(=>x=\dfrac{2}{21}.\dfrac{21}{5}\\ =>x=\dfrac{2}{5}\\ -\dfrac{11}{12}x+\dfrac{15}{2}\left(x+-\dfrac{1}{5}\right)=\dfrac{67}{8}\\ =>-\dfrac{11}{12}x+\dfrac{15}{2}.x-\dfrac{1}{5}=\dfrac{67}{8}\\ =>\left(-\dfrac{11}{12}+\dfrac{15}{2}\right)x=\dfrac{67}{8}+\dfrac{1}{5}\\ =>\left(-\dfrac{11}{12}+\dfrac{90}{12}\right)x=\dfrac{335}{40}+\dfrac{8}{40}\\ =>\dfrac{79}{12}x=\dfrac{343}{40}\\ =>x=\dfrac{343}{40}:\dfrac{79}{12}\\ =>x=\dfrac{343}{40}.\dfrac{12}{79}\\ =>x=\dfrac{343.12}{40.79}\\ =>x=\dfrac{343.3}{10.79}\\ =>x=\dfrac{1029}{790}\)
7 tháng 2 2022
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
PV
0
6 tháng 3 2017
\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+........+\(\frac{1}{x.\left(x+3\right)}\)=\(\frac{101}{1540}\)
3(.\(\frac{1}{5.8}+\frac{1}{8.11}\)+\(\frac{1}{11.14}+.......+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=>\(x+3=308\)
\(x=308-3=305\)
Vậy \(x=305\)
S
6 tháng 3 2017
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 5
x = 303
TN
25 tháng 11 2016
a)\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
b ko hiểu đề
8 tháng 3 2016
Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^
duyệt đi
26 tháng 7 2023
a: =>(x-2)^3*[(x-2)^2-1]=0
=>(x-2)(x-3)(x-1)=0
=>\(x\in\left\{1;2;3\right\}\)
b: =>(x-3)^2*(x-3-1)=0
=>(x-3)(x-4)=0
=>x=3 hoặc x=4
c: =>\(11\cdot\dfrac{6^x}{6}+2\cdot6^x\cdot6=6^{11}\left(11+2\cdot6^2\right)\)
=>6^x(11/6+12)=6^12(11/6+12)
=>x=12
31 tháng 7 2019
\(a;343-4\times\left(x+1\right)=143\)
\(4\times\left(x+1\right)=343-143\)
\(4\times\left(x+1\right)=200\)
\(x+1=200:4\)
\(x+1=50\)
\(x=50-1=49\)
31 tháng 7 2019
Mấy câu hỏi trẻ trâu thì tự tính nhé
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy x=305
đề sai à nhìn số hỗn loạn quá