Question

1/10+1/40+1/88+...+1/(x+2)(x+5)=3/20

Answer 1

làm lại

ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)

=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)

=>\(\frac{1}{x+5}=\frac{1}{20}\)

=>\(x+5=20\)

=>\(x=20-5\)

=>\(x=15\)

Answer 2

ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)

=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)

=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)

=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)

=>\(\frac{1}{x+3}=\frac{1}{20}\)

=>\(x+3=20\)

=>\(x=20-3\)

=>\(x=17\)

Answer 3

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x+2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}:\frac{1}{3}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}\cdot3\\ \frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{1}{2}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{10}{20}\\ \frac{1}{x+5}=\frac{19}{20}\\ \frac{19}{19\left(x+5\right)}=\frac{19}{20}\\ \Rightarrow19\left(x+5\right)=20\\ 19x+95=20\\ 19x=20-95\\ 19x=-75\\ x=\left(-75\right):19\\ x=\frac{75}{19}\)

Vậy \(x=\frac{75}{19}\)

Answer 4

mình viết nhầm đề

Answer 5

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x+2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}:\frac{1}{3}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}\cdot3\\ \frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\\ \frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\\ \frac{1}{x+5}=\frac{10}{20}-\frac{9}{20}\\ \frac{1}{x+5}=\frac{1}{20}\\ \Rightarrow x+5=20\\ x=20-5\\ x=15\)

Vậy x = 15