tìm x:
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
Lưu ý: có cả cách giải:>
\(\dfrac{x+1}{2021}\) + \(\dfrac{x+2}{2020}\) + \(\dfrac{x+1}{2019}\) = -3
Tìm x, biết:
( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2023}\) ) . x = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) + \(\dfrac{2020}{3}\)
+ ... + \(\dfrac{1}{2022}\)
*Thực hiện
1/ (\(\dfrac{2021}{2020}\)+\(\dfrac{2020}{2021}\)) x (\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\))
2/ (\(\dfrac{7}{19}\)-\(\dfrac{5}{12}\)):\(\dfrac{-5}{8}\)-(\(\dfrac{7}{19}\)-\(\dfrac{29}{12}\)):\(\dfrac{5}{8}\)
3/ \(\dfrac{-5}{6}\)x\(\dfrac{7}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{14}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{3}{24}\)
tìm x e Q
a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
Tìm x :
\(\dfrac{x+721}{2020}\) + \(\dfrac{x+21}{700}\) + \(\dfrac{x+721}{2021}\) = -1
a) tìm x :\(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+....+\dfrac{2}{31.344}x=10\)
b)so sánh hai phân số sau : A=\(\dfrac{6^{2020}+1}{6^{2021}+1}\)và B=\(\dfrac{6^{\text{2021}}+1}{\text{6}^{\text{2022}}+1}\)
ét o ét giúp với ạ
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{\dfrac{2020}{1}+\dfrac{2019}{2}+\dfrac{2018}{3}+...+\dfrac{1}{2021}}\)
chi tiết nghen:))
cho A=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2022}\)
B=\(\dfrac{2021}{1}+\dfrac{2020}{2}+\dfrac{2019}{3}+...+\dfrac{1}{2021}\)
tính tỉ số \(\dfrac{B}{A}\)