ta có
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)\(-4\)
\(\Rightarrow\left(\frac{x+1}{212}+1\right)+\left(\frac{x+2}{211}+1\right)+\left(\frac{x+3}{210}+1\right)+\left(\frac{x+4}{209}+1\right)=-4+4\)
=> \(\frac{x+1+212}{212}+\frac{x+2+211}{211}+\frac{x+3+210}{210}+\frac{x+4+209}{209}\) =\(0\)
=> \(\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}\)=\(0\)
=> (x+213) \(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)=0
mà\(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)\(\ne0\)
=>x+213=0 => x=-213
vậy x= -213
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)
\(\Rightarrow\frac{x+1}{212}+1+\frac{x+2}{211}+1+\frac{x+3}{210}+1+\frac{x+4}{209}+1=-4+4=0\)
\(\Rightarrow\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}=0\)
\(\Rightarrow\left(x+213\right)\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)=0\)
\(\Rightarrow x+213=0\Leftrightarrow x=-213\)
