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\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+..........+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2014}{2016}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+.............+\dfrac{2}{x\left(x+1\right)}=\dfrac{2014}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2014}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2014}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2014}{2016}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2016}\)
\(\Leftrightarrow x+1=2016\)
\(\Leftrightarrow x=2015\left(tm\right)\)
Vậy ...........
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
a)\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
(1 - \(\dfrac{1}{2}\)) . (1 - \(\dfrac{1}{3}\)) . (1 - \(\dfrac{1}{4}\)) ... (1 - \(\dfrac{1}{x}\)) = \(\dfrac{1}{2014}\)
=> \(\dfrac{1}{2}\) . \(\dfrac{2}{3}\) . \(\dfrac{3}{4}\) ... \(\dfrac{x-1}{x}\) =\(\dfrac{1}{2014}\)
=> \(\dfrac{1.2.3...x-1}{2.3.4...x}\) = \(\dfrac{1}{2014}\)
=:> \(\dfrac{1}{x}\) = \(\dfrac{1}{2014}\)
Vậy x = 2014
(Mình nghĩa là vậy, có gì sai mong bạn bỏ qua nha)
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
x + 1 + x + 4 + ... + x + 28 = 155
(x + x + x + ... + x) + (1 + 4 + ... + 28) = 155
x . 10 + 145 = 155
x . 10 = 155 - 145
x . 10 = 10
x = 10 : 10
x = 1
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
1a.Vì \(\left|x\right|\) là 1 số tự nhiên nên \(\left|x\right|+2017\ge2017\)(1)
Mà ta đã biết:\(\dfrac{a}{b}\ge\dfrac{a}{b+n}\)với n là một số tự nhiên.
Nên từ (1)suy ra\(\dfrac{2016}{\left|x\right|+2017}\le\dfrac{2016}{2017}\)
Vậy để \(\dfrac{2016}{\left|x\right|+2017}\)lớn nhất thì \(\dfrac{2016}{\left|x\right|+2017}=\dfrac{2016}{2017}\)
1b.Ta thấy:
\(\dfrac{\left|x\right|+2016}{-2017}=\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)
Để \(\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)lớn nhất thì \(-\left(\left|x\right|+2016\right)\)lớn nhất
Mà theo câu a,ta có:\(\left|x\right|\)+2016 là một số tự nhiên nên \(-\left(\left|x\right|+2016\right)\)mang dấu âm hay \(-\left(\left|x\right|+2016\right)\le0\)( chú ý \(-0=0\))
Vậy để \(-\left(\left|x\right|+2016\right)\)lớn nhất hay \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì \(\left|x\right|+2016=0\)
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì nó bằng \(\dfrac{0}{-2017}\)hay nó bằng 0
2)
a)Để \(\dfrac{\left|x\right|+1945}{1975}\)nhỏ nhất thì \(\left|x\right|+1945\) nhỏ nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1945\ge1945\)
\(\Rightarrow\)Để \(\left|x\right|+1945\) nhỏ nhất thì \(\left|x\right|+1945\) = 1945
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+1945}{1975}\)bé nhất thì nó phải bằng \(\dfrac{1945}{1975}\)hay\(\dfrac{389}{395}\)
b)Để \(\dfrac{-1}{\left|x\right|+1}\)thì \(\left|x\right|+1\)bé nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1\ge1\)
\(\Rightarrow\)Để \(\left|x\right|+1\)bé nhất thì \(\left|x\right|+1\)\(=1\)
\(\Rightarrow\)GTNN của \(\dfrac{-1}{\left|x\right|+1}\)là \(\dfrac{-1}{1}\) hay -1
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2014}{2016}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2014}{2016}\)
\(A=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2014}{2016}\)
\(A=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1007}{2016}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)
\(A=\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)
\(A=\dfrac{1}{x+1}=\dfrac{1}{2016}\)\(\Leftrightarrow x+1=2016\Leftrightarrow x=2015\)
ơ cho mình hỏi \(\dfrac{1}{x\cdot\left(x+1\right)}\) không có 2 làm sao tách ra đc