Toán 8 Giải phương trình

Giải phương trình sau :
[tex]\frac{x+2}{2018} + \frac{x+1}{2019} = \frac{x}{2020} + \frac{x-1}{2021}[/tex]

 

Ta có: ...... <=> [tex]\frac{x+2}{2018}+1+\frac{x+1}{2019}+1=\frac{x}{2020}+1+\frac{x-1}{2021}+1[/tex] <=> [tex]\frac{x+2020}{2018}+\frac{x+2020}{2019}=\frac{x+2020}{2020}+\frac{x+2020}{2021}[/tex] <=> [tex](x+2020)(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021})=0[/tex]
Mà [tex]\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\neq 0[/tex] nên [tex]x+2020=0[/tex] <=> x = -2020
Vậy S = { -2020 }

 

PT đã cho tương đương [tex]\frac{x+2}{2018}+1+\frac{x+1}{2019}+1=\frac{x}{2020}+1+\frac{x-1}{2021}+1
\Leftrightarrow \frac{x+2020}{2018}+\frac{x+2020}{2019}=\frac{x+2020}{2020}+\frac{x+2021}{2021}
\Leftrightarrow (x+2020)(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021})=0
=> x=-2020[/tex]

 

[tex]\dpi{50} \left ( \frac{x+2}{2018}+1 \right )+\left ( \frac{x+1}{2019}+1 \right )=\left ( \frac{x}{2020}+1 \right )+\left ( \frac{x-1}{2021}+1 \right )[/tex]
[tex]\dpi{50} \frac{x+2020}{2018}+\frac{x+2020}{2019}=\frac{x+2020}{2020}+\frac{x+2020}{2021}[/tex]
[tex]\dpi{50} \frac{x+2020}{2018}+\frac{x+2020}{2019}-\frac{x+2020}{2020}-\frac{x+2020}{2021}=0[/tex]
[tex]\dpi{50} \left ( x+2020 \right )\cdot \left ( \frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021} \right )=0[/tex]
[tex]\dpi{50} x+2020=0\left ( \frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021} \right )\neq 0[/tex]
x=-2020

 

bạn ơi , hình như bạn viết sai x+2021/2021 rồi , phải là x+2020/2021 chứ nhỉ ?

 

ừ đúng rồi mình gõ nhanh quá nên nhầm chỗ đó