Question

[hồng nguyen thi]

Câu 1 : Tìm X , biết :

    1 / 5 . 8 + 1 / 8 . 11 + ... + 1 / x ( x + 3 ) = 1 / 6 

Answer 1

[Vũ Quang Vinh]

Theo đầu bài ta có:
\(A=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\Rightarrow3A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{1}{6}\)
\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{6}\)
\(x+3=1:\frac{1}{30}\)
\(x=30-3\)
\(x=27\)

Answer 2

[Minh Triều]

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{1}{6}\)

\(\Rightarrow3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=\frac{3}{6}\)

\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\Rightarrow-3.\left(x+3\right)=10\)

=>-3x-9=10

=>-3x=19

=>x=-19/3