Câu 1 :
a) (x + 3)(x 2 + 3x – 5)
= x.(x 2 + 3x – 5) + 3.(x 2 + 3x – 5)
= x.x 2 + x.3x + x.(–5) + 3.x 2 + 3.3x + 3.(–5)
= x 3 + 3x 2 – 5x + 3x 2 + 9x – 15
= x 3 + (3x 2 + 3x 2 ) + (9x – 5x) – 15
= x 3 + 6x 2 + 4x – 15
b) (xy – 1)(xy + 5)
= xy.(xy + 5) + (–1).(xy + 5)
= xy.xy + xy.5 + (–1).xy + (–1).5
= x 2 y 2 + 5xy – xy – 5
= x 2 y 2 + 4xy – 5.
Câu 2 :
a) (x2 – 2x + 1)( x – 1)
= x2.(x – 1) + (–2x).(x – 1) + 1.(x – 1)
= x2.x + x2.(– 1) + (– 2x).x + (–2x).(–1) + 1.x + 1.(–1)
= x3 – x2 – 2x2 + 2x + x – 1
= x3 – (x2 + 2x2) + (2x + x) – 1
= x3 – 3x2 + 3x – 1
b) (x3 – 2x2 + x – 1)(5 – x)
= (x3 – 2x2 + x – 1).5 + (x3 – 2x2 + x – 1).(–x)
= x3.5 + (–2x2).5 + x.5 + (–1).5 + x3.(–x) + (–2x2).(–x) + x.(–x) + (–1).(–x)
= 5x3 – 10x2 + 5x – 5 – x4 + 2x3 – x2 + x
= –x4 + (5x3 + 2x3) – (10x2 + x2) + (5x + x) – 5
= –x4 + 7x3 – 11x2 + 6x – 5
Ta có:
(x3 – 2x2 + x – 1).(x – 5)
= (x3 – 2x2 + x – 1).[–(5 – x)]
= – (x3 – 2x2 + x – 1).(5 – x)
= – (–x4 + 7x3 – 11x2 + 6x – 5)
= x4 – 7x3 + 11x2 – 6x + 5.
