Bài 26 trang 14 sgk toán 8 tập 1


Nội dung bài giảng

Bài 26. Tính:

a) (2x2 + 3y)3;                b) (\(\frac{1}{2}\)x – 3)3

Bài giải:

 a) (2x2 + 3y)3 = (2x2)3  + 3(2x2)2 . 3y + 3 . 2x2 . (3y)2 + (3y)3

                       = 8x6 + 3 . 4x4 . 3y + 3 . 2x2 . 9y2 + 27y3

                        = 8x6 + 36x4y + 54x2y2 + 27y3

b) (\(\frac{1}{2}\)x – 3)3 = \(\left ( \frac{1}{2}x \right )^{3}\)- 3\(\left ( \frac{1}{2}x \right )^{2}\). 3 + 3\(\left ( \frac{1}{2}x \right )\). 32 - 33

                   = \(\frac{1}{8}\)x3 – 3 . \(\frac{1}{4}\)x2 . 3 + 3 . \(\frac{1}{2}\)x . 9 – 27

                   = \(\frac{1}{8}\)x3 – \(\frac{9}{4}\)x2 + \(\frac{27}{2}\)x - 27