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1/2.(1/3+1/6+1/10+...+1/x(x+1))=1/2.2016/2018
1/6+1/12+1/20+...+1/x(x+1)=504/1009
1/2.3+1/3.4+1/4.5+...+1/x(x+1)=504/1009
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=504/1009
1/2-1/x+1=504/1009
x-1/2(x+1)=504/1009
-> 1009(x-1)=504.2(x+1)
1009x-1009=1008x+1008
1009x-1008x=1008+1009
->x=2017
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2016}{2018}\)
\(A=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{x\left(x+1\right):2}\)
\(A=\frac{1}{2\left(2+1\right)}\cdot2+\frac{1}{3\left(3+1\right)}\cdot2+...+\frac{1}{x\left(x+1\right)}.2=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=1-\frac{1}{x+1}=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2016}{2018}=\frac{1}{1009}\)
\(\Rightarrow x+1=1009\Rightarrow x=1008\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4038}\)
\(\Rightarrow x+1=4038\)
\(\Rightarrow x=4037\)
Vậy \(x=4037\)
\(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}+\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\frac{1}{x+1}=\frac{1}{4038}\)
\(x=4037\)
1/2+1/6+1/12+1/20+...+1/x(x+1)=2015/2016
1/1.2+1/2.3+1/3.4+.....+1/x.(x+1)=2015/2016
1-1/2+1/2-1/3+1/3-1/4+......+1/x-1/x+1=2015/2016
1-1/x-1=2015/2016
1/x+1=1-2015/2016
1/x+1=1/2016
=> x+1=2016
x=2016-1
x=2015
vậy x =2015
tích mình nha
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
=>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)
=>\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
=>x+1=2016
=>x=2015
Vậy x=2015
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\cdot\left(x+1\right):2}=\dfrac{2016}{2018}\\ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}:2\\ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2018}\\ \Leftrightarrow x+1=2018\\ x=2018-1\\ x=2017\)
1/6+3x+2=87
3x+2=87-6
3x+2=81
3x+2=34
x+2=4
x =4-2
x =2
2/
(33-3)chia hết cho x =>30 chia hết cho x
(101-11)chia hết cho x 90 chia hết cho x
x thuộc ƯC(30,90)
30=2.3.5
90=2.3.3.5
ƯCLN(30,90)=2.3.5=30
x thuộc ƯC(30,90)=Ư(30)=1 ,2,3,5,6,10,15,30
Sau khi loại các số không hợp điều kiện ta được các số:15,30
Vậy x = 15,30
3/A=2017+20172+20173+.........+20172018
A=(2017+20172)+(20173+20174)+.......(20172017+20172018)
A=2017.(1+2017)+20173.(1+2017)+..........20172017.(1+2017)
A=2017.2018+20173.2018+..................20172017.2018
=>A chia hết cho 2018
cái đề j mk khó hỉu vậy viết đòang hoàng xem nào
Tìm số tự nhiên x biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\cdot\left(x+1\right):2}=\frac{2016}{2018}\)