tìm x :
\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)
K
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TP
1
23 tháng 11 2018
Em chuyển sang cùng một vế rồi ghép cái đầu vói cái thứ 3 cái thứ 2 với cái cuối. :)Dùng quy đồng :)
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
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 \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
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 \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\)
\(\Rightarrow x-2019=0\)
\(\Rightarrow x=2019\)
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
A
4
30 tháng 3 2018
\(\frac{x+4}{2015}+\frac{x+3}{2016}=\frac{x+2}{2017}+\frac{x+1}{2018}\)
\(\Rightarrow\frac{x+4}{2015}+1+\frac{x+3}{2016}+1=\frac{x+2}{2017}+1+\frac{x+1}{2018}+1\)
\(\Rightarrow\frac{x+4+2015}{2015}+\frac{x+3+2016}{2016}=\frac{x+2+2017}{2017}+\frac{x+1+2018}{2018}\)
\(\Rightarrow\frac{x+2019}{2015}+\frac{x+2019}{2016}-\frac{x+2019}{2017}-\frac{x+2019}{2018}=0\)
\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)
=> x + 2019 = 0
=> x = -2019
Vậy x = -2019
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 \(x+2020=0\Rightarrow x=-2020\)
\(\Rightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Rightarrow\orbr{\begin{cases}x=2019\left(1\right)\\\frac{1}{2018}+\frac{1}{2017}=\frac{1}{2016}+\frac{1}{2015}\left(2\right)\end{cases}}\) mà \(\left(2\right)\)không thể xảy ra nên x=2019 là nghiệm của phương trình.