\(\left|x+2\right|=\left|2019x+2020\right|\)
\(\Leftrightarrow x+2=\pm2019x+2020\)
TH1:\(x+2=2019x+2020\)
\(\Leftrightarrow2018x+2018=0\)
\(\Leftrightarrow2018\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
TH2:\(x+2=-\left(2019x+2020\right)\)
\(\Leftrightarrow x+2=-2019x-2020\)
\(\Leftrightarrow-2020x-2022=0\)
\(\Leftrightarrow-2020x=2022\)
\(\Leftrightarrow x=-\frac{2022}{2020}\)
Vậy\(x=-1;-\frac{2022}{2020}\)
Bạn có thể làm bảng xét dấu cũng được,nhưng để chắc ăn hơn thì nên lm cái này,mặc dù hơi dài:v
\(\left|x+2\right|=x+2\Leftrightarrow x+2\ge0\Leftrightarrow x\ge-2\)
\(\left|x+2\right|=-x-2\Leftrightarrow x+2< 0\Leftrightarrow x< -2\)
\(\left|2019x+2020\right|=2019x+2020\Leftrightarrow2019x+2020\ge0\Leftrightarrow2019x\ge-2020\Leftrightarrow x\ge-\frac{2020}{2019}\)
\(\left|2019x+2020\right|=-2019x-2020\Leftrightarrow2019x+2020< 0\Leftrightarrow2019x< -2020\Leftrightarrow x< \frac{-2020}{2019}\)
Xét \(x< -2\)ta có:
\(-x-2=-2019x-2020\)
\(\Leftrightarrow2018x=-2018\)
\(x=-1\left(KTM\right)\)
Xét \(-2\le x< -\frac{2020}{2019}\) ta có:
\(x+2=-2019x-2020\)
\(\Leftrightarrow2020x=-2022\)
\(\Leftrightarrow x=-\frac{1011}{1010}\left(TM\right)\)
Xét \(x\ge-\frac{2020}{2019}\) ta có:
\(x+2=2019x+2020\)
\(\Leftrightarrow-2018x=2018\)
\(\Leftrightarrow x=-1\)
Bọn mik hok cách này nè.Còn cách kia thầy mik ns khi xét trường hợp sẽ thiếu nghiệm đối vs 1 số bài:v
