Nhanh nha,cần gấp lém:
Tìm x:
$\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{10}$+…+$\frac{1}{x(x+1):2}$= $\frac{2021}{2023}$
Nhanh nha,cần gấp lém:
Tìm x:
$\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{10}$+…+$\frac{1}{x(x+1):2}$= $\frac{2021}{2023}$
`1/3+1/6+1/10+…+1/(x(x+1):2)=2021/2023`
`1/(2.3:2)+1/(3.4:2)+1/(4.5:2)+…+1/(x(x+1):2)=2021/2023`
`2/(2.3)+2/(3.4)+2/(4.5)+…+2/(x(x+1))=2021/2023`
`2(1/(2.3)+1/(3.4)+1/(4.5)+…+2/(x(x+1))=2021/2023`
`2(1/2-1/3+1/3-1/4+1/4-1/5+…+1/x+1/(x+1))=2021/2023`
`2(1/2-1(x+1))=2021/2023`
`1/2-1/(x+1)=2021/4046`
`1/(x+1)=1/2-2021/4046`
`1/(x+1)=2/4046`
`1/(x+1)=1/2023`
`x+1=2023`
Suy ra `x=2022`
Vậy `x in{2022}`.
Đáp án: `x=2022`
Giải thích các bước giải:
`1/3 +1/6 + \frac{1}{10}+…+\frac{1}{x(x+1):2}=\frac{2021}{2023}`
`=> 2/6 +\frac{2}{12}+\frac{2}{20}+…+\frac{2}{x(x+1)}=\frac{2021}{2023}`
`=> 2(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+…+\frac{1}{x.(x+1)})=\frac{2021}{2023}`
`=> 2(1/2 -1/3 +1/3 – 1/4 +1/4 -1/5 +…+1/x +\frac{1}{x+1}=\frac{2021}{2023}`
`=>2(1/2 – 0-0-0-…-\frac{1}{x+1})=\frac{2021}{2023}`
`=>2(1/2 -\frac{1}{x+1})=\frac{2021}{2023}`
`=> 1/2 -\frac{1}{x+1}=\frac{2021}{4046}`
`=> \frac{1}{x+1}=1/2-\frac{2021}{4046}`
`=> \frac{1}{x+1} = \frac{1}{2023}`
`=> x+1=2023`
`=> x=2023-1`
`=> x=2022`
Vậy `x=2022`