1/2 + 1/6 + 1/12 + ..... + 1/ x * ( x+1) = 2021/2022
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7 tháng 5 2021
1/2 + 1/6+1/12 + 1/20 +....+ 1/x(x+1) = 2021/2022
1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +...+ 1/x. (x+1) = 2021/2020
1 - 1/2 + 1/2 - 1/3 + 1/3- 1/4 + 1/4 - 1/5 +...+ 1/x - 1/(x+1) = 2021/2020
1 - 1/(x+1) = 2021/2020
1/(x+1) = 1 - 2021/2020
1/(x+1) = -1/2020
1/(x+1) = 1/-2020
x + 1 = - 2020
x = -2020 - 1
x = -2021
8 tháng 5 2021
Giải:
1/2+1/6+1/12+1/20+...+1/x.(x+1)=2021/2022
1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2021/2022
1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=2021/2022
1/1-1/x+1 =2021/2022
1/x+1 =1/1-2021/2022
1/x+1 =1/2022
⇒x+1=2022
x=2022-1
x=2021
Chúc bạn học tốt!
15 tháng 8 2020
\(\frac{1}{2\cdot x}-2021-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{222}=\frac{6}{11}\)
\(\frac{1}{2\cdot x}-2021-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\right)=\frac{6}{11}\)
....
Cái dãy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\) nó không có quy luật, không tính được
XO
15 tháng 8 2020
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
MH
25 tháng 11 2021
\(P=\dfrac{x^2+x+1}{\left(x-1\right)^2}\)
Điều kiện: x≠ \(1\)
Ta có:
\(P=\dfrac{\left(x^2-2x+1\right)+\left(3x-3\right)+3}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^2+3\left(x-1\right)+3}{\left(x-1\right)^2}\)
\(=1+\dfrac{3}{x-1}+\dfrac{3}{\left(x-1\right)^2}\)
\(=3\left[\left(\dfrac{1}{x-1}\right)^2+2.\dfrac{1}{x-1}.\dfrac{1}{2}+\dfrac{1}{4}\right]+\dfrac{1}{4}\)
\(=3\left(\dfrac{1}{x-1}+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\) ≥ \(\dfrac{1}{4}\) (Vì \(3\left(\dfrac{1}{x-1}+\dfrac{1}{2}\right)^2\text{≥}0\) )
Min P=\(\dfrac{1}{4}\) ⇔\(x=-1\)
LM
2
27 tháng 7 2021
Ta có: \(2y+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2020\cdot2021}\right)=\dfrac{4041}{2021}\)
\(\Leftrightarrow2y+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\right)=\dfrac{4041}{2021}\)
\(\Leftrightarrow2y+1-\dfrac{1}{2021}=\dfrac{4041}{2021}\)
\(\Leftrightarrow2y=\dfrac{4041}{2021}+\dfrac{1}{2021}-1\)
\(\Leftrightarrow2y=2-1=1\)
hay \(y=\dfrac{1}{2}\)
OY
1 tháng 8 2021
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\cdots+\frac{1}{x\left(x+1\right)}=\frac{2021}{2022}\)
\(1-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{x}-\frac{1}{x+1}=\frac{2021}{2022}\)
\(1-\frac{1}{x+1}=\frac{2021}{2022}\)
\(\frac{1}{x+1}=1-\frac{2021}{2022}\)
\(\frac{1}{x+1}=\frac{1}{2022}\)
\(x+1=2022\)
\(x=2021\)