Bài 1 (Trang 19 SGK Toán 10, Bộ Cánh diều, Tập 2)

Khai triển các biểu thức sau:

a) (2x + 1)⁴;

b) (3y – 4)⁴;

c) ${(x +\hspace{0.17em}\frac{1}{2})}^4;$

d) ${(x - \frac{1}{3})}^4$.

Hướng dẫn giải

a) (2x + 1)⁴ = (2x)⁴ + 4 . (2x)³ . 1 + 6 . (2x)² . 1² + 4 . (2x) . 1³ + 1⁴

= 16x⁴ + 32x³ + 24x² + 8x + 1.

b) (3y – 4)⁴ = [3y + (– 4)]⁴

= (3y)⁴ + 4 . (3y)³ . (– 4) + 6 . (3y)² . (– 4)² + 4 . (3y) . (– 4)³ + (– 4)⁴

= 81y⁴ – 432y³ + 864y² – 768y + 256.

c) ${(x +\hspace{0.17em}\frac{1}{2})}^4 = x^4 + 4x^3.\frac{1}{2} + 6x^2.{\left(\frac{1}{2}\right)}^2 + 4x.{\left(\frac{1}{2}\right)}^3 + {\left(\frac{1}{2}\right)}^4$

                   $= x^4 + 2x^3 +\hspace{0.17em}\frac{3}{2}{x}^2 +\hspace{0.17em}\frac{1}{2}x +\hspace{0.17em}\frac{1}{16}$

d) ${(x - \frac{1}{3})}^4 = {[x + (-\frac{1}{3}\left)\right]}^4$

$= x^4 + 4x^3.(-\frac{1}{3}) +\hspace{0.17em}6x^2.{(-\frac{1}{3})}^2 + 4x.{(-\frac{1}{3})}^3 + {(-\frac{1}{3})}^4$

$= x^4 -\frac{4}{3}{x}^3 +\hspace{0.17em}\frac{2}{3}{x}^2 - \frac{4}{27}x + \frac{1}{81}$