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TC
1
V
3 tháng 12 2018
\(\Rightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Rightarrow\orbr{\begin{cases}x=2019\left(1\right)\\\frac{1}{2018}+\frac{1}{2017}=\frac{1}{2016}+\frac{1}{2015}\left(2\right)\end{cases}}\) mà \(\left(2\right)\)không thể xảy ra nên x=2019 là nghiệm của phương trình.
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
VT
4
KB
19 tháng 12 2017
Tìm x biết:
\(\frac{x}{2018}+\frac{x+1}{2017}+\frac{x+2}{2016}+\frac{x+3}{2015}=-4\)
Giải:Ta có:\(\frac{x}{2018}+\frac{x+1}{2017}+\frac{x+2}{2016}+\frac{x+3}{2015}=-4\)
\(\Rightarrow\frac{x}{2018}+1+\frac{x+1}{2017}+1+\frac{x+2}{2016}+1+\frac{x+3}{2015}+1=0\)
\(\Rightarrow\frac{x+2018}{2018}+\frac{x+2018}{2017}+\frac{x+2018}{2016}+\frac{x+2018}{2015}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)
\(\Rightarrow x+2018=0\) vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}>0\)
\(\Rightarrow x=-2018\)
Vậy x=-2018 thỏa mãn
19 tháng 12 2017
x2018 +x+12017 +x+22016 +x+32015 =−4
⇒x2018 +1+x+12017 +1+x+22016 +1+x+32015 +1=0
⇒x+20182018 +x+20182017 +x+20182016 +x+20182015 =0
⇒(x+2018)(12018 +12017 +12016 +12015 )=0
⇒x+2018=0 vì 12018 +12017 +12016 +12015 >0
⇒x=−2018
Vậy x=-2018 thỏa mãn
DT
2
MT
11 tháng 7 2015
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)
\(\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
mà \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)nên
\(x+2020=0\)
\(x=-2020\)
11 tháng 7 2015
Cộng 1 vào 2 vế ta có
\(\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)
\(\left(\frac{x+2015}{5}+\frac{5}{5}\right)+\left(\frac{x+2016}{4}+\frac{4}{4}\right)=\left(\frac{x+2017}{3}+\frac{3}{3}\right)+\left(\frac{x+2018}{2}+\frac{2}{2}\right)\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
Vì \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
nên \(x+2020=0\Rightarrow x=-2020\)
TB
3
S
3 tháng 6 2019
\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}=-3\)
\(\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1=0\)
\(\frac{x+2+2018}{2018}+\frac{x+3+2017}{2017}+\frac{x+4+2016}{2016}=0\)
\(\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}=0\)
\(\left(x+2020\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
\(\Rightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
#Sakura
3 tháng 6 2019
\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}=-\overrightarrow{3}\)
=>\(\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1=0\)
=>\(\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}=0\)
=>\(\left(x+2020\right):\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
=>\(\left(x+2020\right)=0\)
=>\(x=0-2020\)
=>\(x=-2020\)
vậy ....
chúc bạn học tốt!
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
TC
1
4 tháng 12 2018
\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)
\(\Rightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Rightarrow\frac{x-1-2018}{2018}+\frac{x-2-2017}{2017}=\frac{x-3-2016}{2016}+\frac{x-4-2015}{2015}\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)
\(\Rightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\)
\(\Rightarrow x-2019=0\)
\(\Rightarrow x=2019\)
H
2
DT
13 tháng 7 2016
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{12\left(x+2015\right)}{60}+\frac{15\left(x+2016\right)}{60}=\frac{20\left(x+2017\right)}{60}+\frac{30\left(x+2018\right)}{60}\)
\(\Rightarrow12x+24180+15x+30240=20x+40340+30x+60540\)
\(\Leftrightarrow-23x=22460\Leftrightarrow x=-\frac{22460}{23}\)
H
1
13 tháng 7 2016
\(\frac{x+2015}{7}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Rightarrow\frac{x+2015}{7}+\frac{7}{7}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Rightarrow\frac{x+2020}{7}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\Rightarrow\frac{x+2020}{7}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{7}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
Mà \(\frac{1}{7}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
\(\Rightarrow x+2020=0\)
\(\Rightarrow x=-2020\)
\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}=-3\)
\(\Leftrightarrow\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1=0\)
\(\Leftrightarrow\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\).
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