4.
a = √(2002•2003•2004•2005+1)
Notez 2002 = x.
a = √[x(x+1)(x+2)(x+3)+1]
a = √[x(x+3)(x+1)(x+2)+1]
a = √[(x²+3x)(x²+3x+2)+1]
a = √[(x²+3x)²+2(x²+3x)+1]
a = √(x²+3x+1)²
a = √(2002²+3•2002+1)²
a = 2002²+3•2002+1 ∈ ℚ
b = ∛(2003•2005³-2004•2002³)
Notez 2003 = x.
x•(x+2)³-(x+1)•(x-1)³ =
= x(x³+6x²+12x+8)-(x+1)(x³-3x²+3x-1)
= x(x³-x³+9x²-9x+9) - x³+3x²-3x+1
= 9x³+9x²+9x - x³+3x²-3x+1
= 8x³+12x²+6x+1
= (2x+1)³
= (2•2003+1)³
= (4007)³
b = ∛(4007)³ = 4007 ∈ ℚ
