n) Theo bài ra ta có: \(\frac{x+1}{2008}=\frac{502}{x+1}\)
=> (x+1).(x+1) = 2008.502
=> (x+1)2 = 1008016
=> (x+1)2 = 10042
=> x+1 = 1004
=> x = 2004-1
=> x = 2003
Vậy x = 2003
p) Theo bà ra ta có: \(\left|\frac{5}{4}.x-\frac{7}{2}\right|-\left|\frac{5}{8}.x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}.x-\frac{7}{2}\right|=\left|\frac{5}{8}.x+\frac{3}{5}\right|\)
=> \(\frac{5}{4}.x-\frac{7}{2}=\pm\left(\frac{5}{8}.x+\frac{3}{5}\right)\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{4}.x-\frac{7}{2}=\frac{5}{8}.x+\frac{3}{5}\\\frac{5}{4}.x-\frac{7}{2}=\frac{-5}{8}.x-\frac{3}{5}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{4}.x-\frac{5}{8}.x=\frac{3}{5}+\frac{7}{2}\\\frac{5}{4}.x+\frac{5}{8}.x=\frac{-3}{5}+\frac{7}{2}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{8}.x=\frac{41}{10}\\\frac{15}{8}.x=\frac{29}{10}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}x=\frac{164}{25}\\x=\frac{116}{75}\end{array}\right.\)
Vậy x=\(\frac{164}{25}\) hoặc x=\(\frac{116}{75}\)
