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\(\frac{1}{2\cdot x}-2021-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{222}=\frac{6}{11}\)
\(\frac{1}{2\cdot x}-2021-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\right)=\frac{6}{11}\)
....
Cái dãy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\) nó không có quy luật, không tính được
Sửa đề\(\frac{1}{2x-2021}-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{220}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{10}-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}.\frac{10}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{5}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}=1\)
=> 2x - 2021 = 1
=> 2x = 2022
=> x = 1011
Vậy x = 1011
\(x\times2019-x=2018\times2018+2018\)
\(x\times\left(2019-1\right)=2018\times\left(2018+1\right)\)
\(x\times2018=2018\times2019\)
\(\Rightarrow x=2019\)
X x 2019 - x =2018 x 2018 + 2018
X x(2019-1) = 2018 x(2018 + 1)
X x 2018 = 2018 x 2019
=> X=2019
Vậy X=2019.
A = (1+1) \(\times\) (1+\(\dfrac{1}{2}\)) \(\times\) (1+\(\dfrac{1}{3}\)) \(\times\)....\(\times\)(1+ \(\dfrac{1}{2020}\))
A = 2 \(\times\) \(\dfrac{3}{2}\) \(\times\) \(\dfrac{4}{3}\) \(\times\) \(\dfrac{5}{4}\) \(\times\).......\(\times\) \(\dfrac{2021}{2020}\)
A =\(\dfrac{2\times3\times4\times....\times2020}{2\times3\times4\times....\times2020}\) \(\times\) \(\dfrac{2021}{1}\)
A = 1 \(\times\) 2021
A = 2021
A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+..+9\right)}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times\left(1+1\right)}{2}+\frac{2\times\left(2+1\right)}{2}+\frac{3\times\left(3+1\right)}{2}...+\frac{9\times\left(9+1\right)}{2}}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times2}{2}+\frac{2\times3}{2}+\frac{3\times4}{2}+...+\frac{9\times10}{2}}{1\times2+2\times3+3\times4+...+9\times10}\)
\(=\frac{\frac{1}{2}\times\left(1\times2+2\times3+3\times4+...+9\times10\right)}{1\times2+2\times3+3\times4+...+9\times10}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\)
1/4.5+1/5.6+.+1/x(x+1)=21/100
=>1/4-1/5+1/5-1/6+.......+1/x-1/(x+1)=21/100
=>1/4-1/(x+1)=21/100
=>1/(x+1)=1/4-21/100
=>1/(x+1)=1/25
=>x+1=25
=>x=24
vậy.......
c,Dãy trên có số số hạng là:
(52-2):2+1=26 số hạng
Tổng của dãy trên là:
(52+2).26:2=702
Đáp số :.....
\(\left(x-2020\right)\left(x-2021\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2020=0\\x-2021=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0+2020=2020\\x=0+2021=2021\end{cases}}}\)
Vậy \(x=2020;2021\)
\(\left(x-2020\right)\left(x-2021\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-2020=0\\x-2021=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2020\\x=2021\end{cases}}\)
vậy x=2020 hoặc x=2021