Tìm cặp số nguyên ( x , y ) thỏa mãn :
a, | 2x + 3 | + | 2x – 1 | = $\frac{8}{2(y-5)^{2}+2}$
b, | x + 3 | + | x – 1 | = $\frac{16}{| y – 2 | + | y + 2 |}$
c, | x – 2y – 1 | + 5 = $\frac{10}{| y – 4 | + 2}$
d, ( x – 2 ) $(x +2 )^{8}$ + 4 = $\frac{20}{3|y+2|+5}$
e, 2|x – 2007 | + 3 = $\frac{6}{| y – 2008|+2}$
