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TH
1
TT
22 tháng 4 2019
Sửa đề: \(\frac{x-3}{2018}\rightarrow\frac{x-3}{2016}\)
\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-2019=0\) (Vì \(\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)\ne0\) )
\(\Leftrightarrow x=2019\)
Vậy \(S=\left\{2019\right\}\)
18 tháng 8 2019
Này Thục Trinh, chỗ mà \(\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
bạn phải có đóng mở ngoặc vào chứ
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
27 tháng 12 2017
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
DT
2
16 tháng 9 2018
\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2017}{8}\)
\(\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
mà \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
Vậy x = 2017
16 tháng 9 2018
\(\Rightarrow\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\Rightarrow\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Rightarrow x-2017=0\)(vì \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\))
=>x=2017
Chúc bạn học tốt
8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
LL
2
XO
3 tháng 9 2020
ĐK \(2018x\ge0\Rightarrow x\ge0\)
Khi đó \(x+\frac{1}{2018}\ge0;x+\frac{2}{2018}\ge0;...;x+\frac{2017}{2018}\ge0\)
Ta có \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)(Vế trái có 2017 hạng tử)
<=> \(x+\frac{1}{2018}+x+\frac{2}{2018}+...+x+\frac{2017}{2018}=2018x\)
<=> \(\left(x+x+...x\right)+\left(\frac{1}{2018}+\frac{2}{2018}+...+\frac{2017}{2018}\right)=2018x\)
2017 hạng tử x 2017 số hạng
=> \(2017x+\frac{1+2+...+2017}{2018}=2018x\)
=> \(x=\frac{2017.\left(2017+1\right):2}{2018}\)
\(\Rightarrow x=\frac{2017}{2}=1008,5\)(tm)
Vậy x = 1008,5
3 tháng 9 2020
Vì \(\left|x+\frac{1}{2018}\right|\ge0\forall x\)
\(\left|x+\frac{2}{2018}\right|\ge0\forall x\)
\(\left|x+\frac{3}{2018}\right|\ge0\forall x\)
.......................................
\(\left|x+\frac{2017}{2018}\right|\ge0\forall x\)
\(\Rightarrow\)\(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|\ge0\forall x\)
mà \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)
\(\Rightarrow\)\(2018x\ge0\forall x\)\(\Rightarrow\)\(x\ge0\)
\(\Rightarrow\)\(x+\frac{1}{2018}+x+\frac{2}{2018}+x+\frac{3}{2018}+...+x+\frac{2017}{2018}=2018x\)
\(\Leftrightarrow\)\(2017x+\frac{1}{2018}+\frac{2}{2018}+\frac{3}{2018}+...+\frac{2017}{2018}=2018x\)
\(\Leftrightarrow\)\(\frac{1+2+3+...+2017}{2018}=x\)
\(\Leftrightarrow\)\(x=\frac{\left[\left(2017+1\right).2017\right]:2}{2018}\)
\(\Leftrightarrow\)\(x=\frac{2035153}{2018}\)
\(\Leftrightarrow\)\(x=\frac{2017}{2}=1008,5\)
Vậy \(x=1008,5\)
bạn cần nói rõ đề hơn nhé
3 dấu giá trị tuyệt đối là sao