Bài 1 trang 5 sgk toán 8 tập 1.


Nội dung bài giảng

1. Làm tính nhân:

a) x2(5x3 – x - \(\frac{1}{2}\));

b) (3xy – x2 + y) \(\frac{2}{3}\)x2y;

c) (4x3– 5xy + 2x)(- \(\frac{1}{2}\)xy).

Bài giải:

a) x2(5x3 – x - \(\frac{1}{2}\)) = x2. 5x3 + x2 . (-x) + x2 . (-\(\frac{1}{2}\))

                           = 5x5 – x3 – \(\frac{1}{2}\)x2

b) (3xy – x2 + y) \(\frac{2}{3}\)x2y = \(\frac{2}{3}\)x2y . 3xy + \(\frac{2}{3}\)x2y . (- x2) + \(\frac{2}{3}\)x2y . y

                                  = 2x3y2 – \(\frac{2}{3}\)x4y + \(\frac{2}{3}\)x2y2

c) (4x3– 5xy + 2x)(- \(\frac{1}{2}\)xy) = - \(\frac{1}{2}\)xy . 4x3 + (- \(\frac{1}{2}\)xy) . (-5xy) + (- \(\frac{1}{2}\)xy) . 2x

                                      = -2x4y + \(\frac{5}{2}\)x2y2 - x2y.