Toán tìm x biết 1/3+1/6+1/10+……+1/ x(x+1):2=2019/2021 08/10/2021 By Claire tìm x biết 1/3+1/6+1/10+……+1/ x(x+1):2=2019/2021
1/3 + 1/6 + 1/10 +…+ 1/x(x+1):2 = 2019/2021 2.(1/6 + 1/12 + 1/20 +…+ 1/2x.(x+1):2) = 2019/2021 2.(1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5+…+ 1/x – 1/x+1) = 2019/2021 2.(1/2 – 1/x+1) = 2019/2021 1/2 – 1/x+1 = 2019/2021 : 2 1/2 – 1/x+1 = 2019/4042 1/x+1 = 1/2 – 2019/2042 1/x+1 = 2021/4042 – 2019/4042 1/x+1 = 2/4042 1/x+1 = 1/2021 => x+1 = 2021 x = 2021 – 1 x = 2020 Trả lời
Chia 2 vào 2 vế ta có $\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + \cdots + \dfrac{1}{x(x+1)} = \dfrac{2019}{4042}$ Tách ra ta có $\dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5} + \cdots + \dfrac{1}{x(x+1)} = \dfrac{2019}{4042}$ Tách ra $\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + \cdots + \dfrac{1}{x} – \dfrac{1}{x+1} = \dfrac{2019}{4042}$ Rút gọn $\dfrac{1}{2} – \dfrac{1}{x+1} = \dfrac{2019}{4042}$ Chuyển vế ta có $\dfrac{1}{x+1} = \dfrac{1}{2} – \dfrac{2019}{4042}$ Quy đồng $\dfrac{1}{x+1} = \dfrac{1}{2021}$ Suy ra $x + 1 = 2021$ Vậy $x = 2020$ Nghiệm là $x = 2020$. Trả lời
1/3 + 1/6 + 1/10 +…+ 1/x(x+1):2 = 2019/2021
2.(1/6 + 1/12 + 1/20 +…+ 1/2x.(x+1):2) = 2019/2021
2.(1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5+…+ 1/x – 1/x+1) = 2019/2021
2.(1/2 – 1/x+1) = 2019/2021
1/2 – 1/x+1 = 2019/2021 : 2
1/2 – 1/x+1 = 2019/4042
1/x+1 = 1/2 – 2019/2042
1/x+1 = 2021/4042 – 2019/4042
1/x+1 = 2/4042
1/x+1 = 1/2021
=> x+1 = 2021
x = 2021 – 1
x = 2020
Chia 2 vào 2 vế ta có
$\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + \cdots + \dfrac{1}{x(x+1)} = \dfrac{2019}{4042}$
Tách ra ta có
$\dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5} + \cdots + \dfrac{1}{x(x+1)} = \dfrac{2019}{4042}$
Tách ra
$\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + \cdots + \dfrac{1}{x} – \dfrac{1}{x+1} = \dfrac{2019}{4042}$
Rút gọn
$\dfrac{1}{2} – \dfrac{1}{x+1} = \dfrac{2019}{4042}$
Chuyển vế ta có
$\dfrac{1}{x+1} = \dfrac{1}{2} – \dfrac{2019}{4042}$
Quy đồng
$\dfrac{1}{x+1} = \dfrac{1}{2021}$
Suy ra
$x + 1 = 2021$
Vậy
$x = 2020$
Nghiệm là $x = 2020$.