Question

1 / giải phương trình sau:

\(\frac{1}{\left(x+2000\right).\left(x+2001\right)}+\frac{1}{\left(x+2001\right).\left(x+2002\right)}...\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

Answer 1

\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)

=> x = -1999 hoặc x = - 2008