Tìm x biết
\(\frac{x-1}{2019}\)+ \(\frac{x-2}{2018}\)- \(\frac{x-3}{2017}\)= \(\frac{x-4}{2016}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
7 tháng 8 2018
\(a)\) 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 ~
L
1
24 tháng 8 2019
Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))
\(\Leftrightarrow x=1\)
Vạy x=1
DT
3
22 tháng 7 2018
\(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
\(\Rightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)
\(\Rightarrow\frac{x+4+2016}{2016}+\frac{x+3+2017}{2017}=\frac{x+2+2018}{2018}+\frac{x+1+2019}{2019}\)
\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}=\frac{x+2020}{2018}+\frac{x+2020}{2019}\)
\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
\(\Rightarrow x+2020=0\) vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x=-2020\)
DN
1
VM
17 tháng 8 2019
\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}=\frac{x-2020}{2017}+\frac{x-2020}{2016}\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x-2020=0\)
\(\Leftrightarrow x=0+2020\)
\(\Rightarrow x=2020\)
Vậy \(x=2020.\)
Chúc bạn học tốt!
CM
1
16 tháng 7 2018
<=>[ (x-1)/2019] -1 +[(x-2)/2018]-1 = [(x-3)/2017]-1 +[(x-4)/2016] -1
<=> (x-2020)/2019 +(x-2020)/2018 = (x-2020)/2017 + (x-2020)/2016
<=> (x-2020)( 1/2019+1/2018-1/2017-1/2016)= 0
=> x-2020= 0 => x= 2020
12 tháng 1 2020
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}-1\right)+\left(\frac{x+2}{2018}-1\right)=\left(\frac{x+3}{2017}-1\right)+\left(\frac{x+4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}=\frac{x+2020}{2017}+\frac{x+2020}{2016}\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2020=0:\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)\)
\(\Leftrightarrow x+2020=0\)
Còn lại tự làm :V
12 tháng 1 2020
Lộn chỗ này , thay chút nha !
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)
Sorry =))
R
2 tháng 8 2019
a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Leftrightarrow\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}>0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
\(\Leftrightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)
\(\Leftrightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
2 tháng 8 2019
a) \(\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
=>\(x+2=0\)
=>\(x=-2\)
nếu có sai thì mong bn thông cảm nha
TS
4
19 tháng 6 2017
Ta có : \(\frac{x-1}{2017}+\frac{x-2}{2018}-\frac{x-3}{2019}=\frac{x-4}{2020}\)
\(\Rightarrow\frac{x-1}{2017}+\frac{x-2}{2018}=\frac{x-4}{2020}+\frac{x-3}{2019}\)
\(\Rightarrow1+\frac{x-1}{2017}+1+\frac{x-2}{2018}=1+\frac{x-4}{2020}+1+\frac{x-3}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}=\frac{2016+x}{2020}+\frac{2016+x}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}-\frac{2016+x}{2019}-\frac{2016+x}{2020}=0\)
\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\text{Mà :
}\)\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)
\(\text{Nên : }\) \(2016+x=0\)
\(\Rightarrow x=-2016\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)
\(\Leftrightarrow\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}-\frac{x-4}{2016}=0\)
\(\Leftrightarrow\frac{x-1}{2019}-1+\frac{x-2}{2018}-1-\frac{x-3}{2017}+1-\frac{x-4}{2016}+1=0\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}-2=\frac{x-3}{2017}+\frac{x-4}{2016}-2\)
\(\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\frac{x-1-2019}{2019}+\frac{x-2-2018}{2018}=\frac{x-3-2017}{2017}+\frac{x-4-2016}{2016}\)
\(\frac{x-2020}{2019}+\frac{x-2020}{2018}=\frac{x-2020}{2017}+\frac{x-2020}{2016}\)
\(\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Rightarrow x-2020=0\)
Vậy \(x=2020\)