\(\frac{x-3}{2015}+\frac{x-2}{2016}=\frac{x-2016}{2}+\frac{x-2015}{3}\)
\(\Leftrightarrow\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-2}{2016}-1\right)=\left(\frac{x-2016}{2}-1\right)+\left(\frac{x-2015}{3}-1\right)\)
\(\frac{x-2018}{2015}+\frac{x-2018}{2016}-\frac{x-2018}{2}-\frac{x-2018}{3}=0\)
\(\Leftrightarrow\left(x-2018\right).\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}< 0\)
nên x - 2018 = 0
,<=> x = 2018
Vậy phương có 1 nghiệm là x = 2018
pt <=> (x-3/2015 - 1) + (x-2/2016 - 1) = (x-2016/2 - 1) + (x-2015/3 - 1)
<=> x-2018/2015 + x-2018/2016 = x-2018/2 + x-2018/3
<=> x-2018/2 + x-2018/3 - x-2018/2015 - x-2018/2016 = 0
<=> (x-2018).(1/2+1/3-1/2015-1/2016) = 0
<=> x-2018 = 0 ( vì 1/2+1/3-1/2015-1/2016 > 0 )
<=> x=2018
Tk mk nha