Question

[nguyen truong giang]

1/15.1/35.1/63.1/99.1/143 = ?

Answer 1

[Hoàng Đình Vinh]

ok !!!

giải :

ta có \(\frac{1}{15}\times\frac{1}{35}\times\frac{1}{63}\times\frac{1}{99}\times\frac{1}{143}\)

      =\(\frac{1}{3\times5}\times\frac{1}{5\times7}\times\frac{1}{5\times7}\times\frac{1}{7\times9}\times\frac{1}{9\times11}\times\frac{1}{11\times13}\)

      =2(\(\frac{1}{3\times5}\times\frac{1}{5\times7}\times\frac{1}{5\times7}\times\frac{1}{7\times9}\times\frac{1}{9\times11}\times\frac{1}{11\times13}\))

      =\(\frac{2}{3\times5}\times\frac{2}{5\times7}\times\frac{2}{7\times9}\times\frac{2}{9\times11}\times\frac{2}{11\times13}\)

      =\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

      =\(\frac{1}{3}-\frac{1}{13}\)

      =\(\frac{10}{39}\)

**** cho mình nha nguyen truong giang ^^mình làm đúng đấy !!!!!!!

Answer 2

[Đỗ Lê Tú Linh]

\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)

\(=2\cdot\frac{1}{2}\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\cdot\left[\left(\frac{1}{3}-\frac{1}{13}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{11}-\frac{1}{11}\right)\right]\)

\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}\cdot\left(\frac{13}{39}-\frac{3}{39}\right)=\frac{1}{2}\cdot\frac{10}{39}=1\cdot\frac{5}{39}=\frac{5}{39}\)