( x2 + 8x + 7 ) ( x2 + 8x + 15 ) + 15
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
đặt t = x2 + 8x + 7
có : t ( t + 8 ) + 15
= t2 + 8t + 15
= t2 + 5t + 3t + 15
= ( t2 + 5t ) + ( 3t + 15 )
= t ( t + 5 ) + 3 ( t + 5 )
= ( t + 3 ) ( t + 5 )
mà t = x2 + 8x + 7
⇒ ( x2 + 8x + 10 ) ( x2 + 8x + 12 )