Question

[Nguyễn Thanh Hằng]

Tìm x biết rằng :

a, \(\frac{1}{5.8}\) + \(\frac{1}{8.11}\) + \(\frac{1}{11.14}\) +............+ \(\frac{1}{x\left(x+3\right)}\) = \(\frac{101}{1540}\)

b, 1 + \(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{10}\) +........+ \(\frac{1}{x\left(x+1\right):2}\) = \(1\frac{1991}{1993}\)

Giải chi tiết giúp mk nha các bn!! mk cảm ơn trước!!

Answer 1

[Nguyễn Huy Tú]

a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

Vậy x = 305

Answer 2

[Phan Hữu Bảo Linh]

a, \(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+...+\(\dfrac{1}{x\left(x+3\right)}\)=\(\dfrac{101}{1540}\)

\(\dfrac{1}{3}\)(\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)+\(\dfrac{3}{11.14}\)+...+\(\dfrac{3}{x\left(x+3\right)}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\)=\(\dfrac{101}{1540}\) : \(\dfrac{1}{3}\)

\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\)=\(\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}\)=\(\dfrac{1}{5}\)-\(\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}\)=\(\dfrac{1}{308}\)

<=>1(x+3)=308.1

<=>1(x+3)=308

<=> x+3=308:1

<=> x+3=308

<=> x=308-3

<=> x=305

b,1+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{1}{x\left(x+1\right):2}\)=1\(\dfrac{1991}{1993}\)

\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{3984}{1993}\)\(2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3984}{1993}\)

\(2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)

\(2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)

\(1-\dfrac{1}{x+1}=\dfrac{3984}{1993}:2\)

\(1-\dfrac{1}{x+1}=\dfrac{1992}{1993}\)

\(\dfrac{1}{x+1}=1-\dfrac{1992}{1993}\)

\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)

<=>1(x+1)=1993.1

<=>1(x+1)=1993

<=> x+1=1993 : 1

<=> x+1=1993

<=> x=1993-1

<=> x=1992