1+1/3+1/6+1/10+.......+1/x(x+1):2=2018/2019
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4038}\)
\(\Rightarrow x+1=4038\)
\(\Rightarrow x=4037\)
Vậy \(x=4037\)
\(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}+\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\frac{1}{x+1}=\frac{1}{4038}\)
\(x=4037\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x :
a) (x - 3) + (x - 2) + (x - 1) + .... + 10 + 11 = 11
(x - 3) + (x - 2) + (x - 1) + .... + 10 = 0
[(x - 3) + (x - 2) + (x - 1)] + (0 + 1 + 2 + ... + 10) = 0
[(x - 3) + (x - 2) + (x - 1)] + 55 = 0
x - 3 + x - 2 + x - 1 = -55
x + x + x - (3 + 2 + 1) = -55
x3 - 6 = -55
x3 = -55 + 6
x3 = -49
x = -49 : 3
x = -\(\frac{49}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x-3) + (x-2) + ( x-1) + ..... + 10 + 11 = 11
(x-3) + (x-2) + ( x-1) + ..... + 10 = 0
Gọi số các số hạng từ x-3 đến 10 là n
Ta có : [10 + (x-3)].n : 2 = 0
(x+7).n = 0
Vì n ≠ 0 ( n là số các số hạng )
Nên x+7 = 0
x = 0-7
x = -7
Vậy x = -7
b)
x + ( x + 1 ) + ( x + 2 ) + ... + 2018 + 2019 = 2019
⇒ x + ( x +1 ) + ... + 2018 = 0
⇒ x + ( x + 1 ) + ... + ( x + 2018 ) = 1 + 2 + ... + 2018
⇒ x = 0
vậy x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times0\)
\(A=0\)
\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x.\left(x+1\right)}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(1-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019.2}\)
Tự làm nốt