Ta có: VT = \(\dfrac{x+1}{2021}\)+1 - (\(\dfrac{x+2}{2020}\)+1) = \(\dfrac{x+3}{2019}\)+1=VP
=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}-\dfrac{x+2022}{2019}=0\)
=>\(\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}\right)=0\)
=>x +2022 = 0=> x =-2022
1) giải phương trình :
a) 3.(2x-3)=5x+1
b) \(\dfrac{x+1}{2021}\)+\(\dfrac{x+2}{2020}\)+\(\dfrac{x+3}{2019}\)+\(\dfrac{x+2023}{2}\)=0
giải chi tiết giúp mik vs ah
Giải phương trình sau: \(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
Cho \(\dfrac{x}{2020}+\dfrac{y}{2021}+\dfrac{z}{2022}=1\) và \(\dfrac{2020}{x}+\dfrac{2021}{y}+\dfrac{2022}{z}=0\) \(\left(x,y,z\ne0\right)\)
Chứng minh rằng \(\dfrac{x^2}{2020^2}+\dfrac{y^2}{2021^2}+\dfrac{z^2}{2022^2}=1\)
Giải các pt sau:
1)\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+1}=\dfrac{3}{2-x}\)
2)\(\dfrac{3x+1}{1-3x}+\dfrac{3+x}{3-x}=2\)
3)\(\dfrac{8x-2}{3}=1+\dfrac{5-2x}{4}\)
4)
\(\dfrac{x}{x+1}-\dfrac{2x+3}{x}=\dfrac{-3}{x+1}-\dfrac{3}{x}\)
5)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
6)\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
giúp mình với cám ơn
Giải PT sau:
\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-\dfrac{x}{4}\)
Giải PT: \(\dfrac{2x-1}{3}\) - \(\dfrac{x-1}{2}\) + \(\dfrac{x+1}{6}\) = 1
Mn ơi giúp e bài này với ạ, e cần gấp lắm. E sắp thi cuối năm r ạ hmu-
\(\dfrac{x+2}{2019}+\dfrac{x+3}{2018}=\dfrac{x+4}{2017}+\dfrac{x}{2021}\)
E cảm ơn mn nhìu lắm!!! Mọng mn giải chi tiết cho e hiểu ạ hyhy XĐ
\(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}=\dfrac{2}{-x^2+6x-8}\)
Giải pt
giải pt
\(\dfrac{x-2}{x+1}=\dfrac{3+x}{x-1}\)
