Chứng minh rằng x^2002 + x^2000 + 1 chia hết cho x^2 + x + 1.
Chứng minh rằng x2002 + x2000 + 1 chia hết cho x2 + x + 1.
Ta có:
A = x2002 + x2000 + 1 = x2002 + x2001 + x2000 − (x2001 − 1)
= x2000(x2 + x + 1) − [(x3)667 − 1]
= x2000(x2 + x + 1) − (x3 − 1)[(x3)666 + (x3)665 + … + 1]
= x2000(x2 + x + 1) − (x − 1)(x2 + x + 1)[(x3)666 + (x3)665 + … + 1]
= (x2 + x + 1){x2000 − (x − 1)[(x3)666 + (x3)665 + … + 1]}.
Vây x2002 + x2000 + 1 chia hết cho x2 + x + 1.