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1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> \(\dfrac{2}{2}\) + \(\dfrac{2}{6}\) + \(\dfrac{2}{12}\) + \(\dfrac{2}{20}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> \(\dfrac{1.2}{1.2}\) + \(\dfrac{1.2}{2.3}\) + \(\dfrac{1.2}{3.4}\) + \(\dfrac{1.2}{4.5}\) + ... + \(\dfrac{1.2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> 2(\(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{x\left(x+1\right)}\)) = 1\(\dfrac{2015}{2017}\)
=> 2(1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\)) = 1\(\dfrac{2015}{2017}\)
=> 2(1 - \(\dfrac{1}{x+1}\)) = \(\dfrac{4032}{2017}\)
=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{4032}{2017}\) : 2
=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{2016}{2017}\)
=> \(\dfrac{1}{x+1}\) = 1 - \(\dfrac{2016}{2017}\)
=> \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2017}\)
=> x + 1 = 2017
=> x = 2017 - 1
=> x = 2016
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 \(x=14\).
\(\dfrac{1}{3}+\dfrac{1}{6}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow2.\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2017}{4038}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2019}\\ \Rightarrow x=2018\)
Đặt \(S=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1008}\right)\)
\(=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\)
Nên:
\(A=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)\(=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)\(\Rightarrow A=1\)
Vậy A = 1
Chúc bạn học tốt!!
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)
=> \(x=56\)
2) => 18x = 18
=> x = 1
3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)
=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)
=> \(x=\dfrac{-1}{2}\)
4) 45%.x =\(\dfrac{3}{5}\)
=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)
=> \(x=\dfrac{4}{3}\)
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) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
Câu 1:
a, \(\left|-5\right|=5\)
b, \(\left|10\right|=10\)
c, \(\left|-5\right|-\left|10\right|=5-10=-5\)
d, -15.30= -450
Câu 2:
a, Ta có: \(\dfrac{10}{21}.\dfrac{14}{25}=\dfrac{10.14}{21.25}=\dfrac{5.2.7.2}{3.7.5.5}=\dfrac{2.2}{3.5}=\dfrac{4}{15}\)
c, Ta có: \(-\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{-5.2+3.3}{12}=\dfrac{-10+9}{12}=\dfrac{-1}{12}\)
d, \(\dfrac{11}{17}.\dfrac{3}{2017}+\dfrac{11}{17}.\dfrac{2014}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}\left(\dfrac{3}{2017}+\dfrac{2014}{2017}\right)-1\dfrac{11}{17}\)
\(=\dfrac{11}{17}.\dfrac{2017}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}-1-\dfrac{11}{17}=-1\)
Câu 7: a, Để A có nghĩa khi \(x+2\ne0\) \(\Leftrightarrow x=-2\)
b, Ta có: \(A=2\)
<=> \(\dfrac{x-1}{x+2}=2\)
<=> \(\dfrac{x-1}{x+2}-2=0\)
<=> \(\dfrac{x-1}{x+2}-\dfrac{2x+4}{x+2}=0\)
<=> \(\dfrac{x-1-2x-4}{x+2}=0\)
<=> \(\dfrac{-x-5}{x+2}=0\)
<=> -x-5=0
<=> -x=5
<=> x= -5
\(\Leftrightarrow2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2015}{2017}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2015}{4034}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2015}{4034}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
=>x+1=2017
hay x=2016