\(\frac{x-2}{2016}+\frac{x-3}{2017}+\frac{x-4}{2018}+3=0\\ \Leftrightarrow\left(\frac{x-2}{2016}+1\right)+\left(\frac{x-3}{2017}+1\right)+\left(\frac{x-4}{2018}+1\right)=0\\ \Leftrightarrow\frac{x+2014}{2016}+\frac{x+2014}{2017}+\frac{x+2014}{2018}=0\\ \Leftrightarrow\left(x+2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\\ Vì\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\ne0\\ \Rightarrow x+2014=0\\ \Leftrightarrow x=-2014\\ Vậy...\)

\(\frac{x-2}{2016}+1+\frac{x-3}{2017}+1+\frac{x-4}{2018}+1+3-3=0\)

\(\frac{x-2014}{2016}+\frac{x-2014}{2017}+\frac{x-2014}{2018}=0\)

\(\left(x-2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\)

⇒x-2014=0➞x=2014

Ta có\(\frac{x-2}{2016}+\frac{x-3}{2017}+\frac{x-4}{2018}+3=0\)

\(\Leftrightarrow\frac{x-2}{2016}+1+\frac{x-3}{2017}+1+\frac{x-4}{2018}=0\)

\(\Leftrightarrow\frac{x+2014}{2016}+\frac{x+2014}{2017}+\frac{x+2014}{2018}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\) Vì \(\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)>0\)

\(\Rightarrow x+2014=0\)

\(\Rightarrow x=-2014\)

a)x=0

a) \(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}=0\)

\(\frac{77x}{60}=0\)

\(77x=0.60\)

\(77x=0\)

\(x=0\)

\(\frac{a+1}{2019}+\frac{a+2}{2018}=\frac{a+3}{2017}+\frac{a+4}{2016}\Leftrightarrow\frac{a+2020}{2019}+\frac{a+2020}{2018}=\frac{a+2020}{2017}+\frac{a+2020}{2016}\) 

\(\left(a+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\Rightarrow a+2020=0\Leftrightarrow a=-2020\)

Bài này có quy tắc đấy

Mình sẽ gúp cậu nhưng sai thì thôi nhé

BÀi này mình chỉ nói kết quả thôi , là : 1/2018

Quy tắc là :

Lấy số đầu + số cuối : cho số khoảng cách + 1

x+2015/5 + x+2016/4=x+2017/3 + x+2018/2

\(\Rightarrow\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)

\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)

\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Rightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Rightarrow x+2020=0\).Do \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)

\(\Rightarrow x=-2020\)