Nội dung bài giảng
44. Phân tích các đa thức sau thành nhân tử:
a) x3 + \(\frac{1}{27}\); b) (a + b)3 – (a – b)3
c) (a + b)3 + (a – b)3 ; d) 8x3 + 12x2y + 6xy2 + y3
e) - x3 + 9x2 – 27x + 27.
Bài giải:
a) x3 + \(\frac{1}{27}\) = x3 + (\(\frac{1}{3}\))3 = (x + \(\frac{1}{3}\))(x2 – x . \(\frac{1}{3}\)+ (\(\frac{1}{3}\))2)
=(x + \(\frac{1}{3}\))(x2 – \(\frac{1}{3}\)x + \(\frac{1}{9}\))
b) (a + b)3 – (a - b)3
= [(a + b) – (a – b)][(a + b)2 + (a + b) . (a – b) + (a – b)2]
= (a + b – a + b)(a2 + 2ab + b2 + a2 – b2 + a2 – 2ab + b2)
= 2b . (3a3 + b2)
c) (a + b)3 + (a – b)3 = [(a + b) + (a – b)][(a + b)2 – (a + b)(a – b) + (a – b)2]
= (a + b + a – b)(a2 + 2ab + b2 – a2 +b2 + a2 – 2ab + b2]
= 2a . (a2 + 3b2)
d) 8x3 + 12x2y + 6xy2 + y3 = (2x)3 + 3 . (2x)2 . y +3 . 2x . y + y3 = (2x + y)3
e) - x3 + 9x2 – 27x + 27 = 27 – 27x + 9x2 – x3 = 33 – 3 . 32 . x + 3 . 3 . x2 – x3 = (3 – x)3