Giải bài 1 trang 19

Đề bài

Khai triển các biểu thức sau:

a) ${\left(2x+1\right)}^4$;

b) ${\left(3y-4\right)}^4$;       

c) ${\left(x+\frac{1}{2}\right)}^4$;   

d) ${\left(x-\frac{1}{3}\right)}^4$.

Lời giải chi tiết

a) ${\left(2x+1\right)}^4$

$={\left(2x\right)}^4+4.{\left(2x\right)}^3{.1}^1+6.{\left(2x\right)}^2{.1}^2+4.\left(2x\right){.1}^3+1^4$

$=16x^4+32x^3+24x^2+8x+1$.

b) ${\left(3y-4\right)}^4={\left[3y+\left(-4\right)\right]}^4$

$={\left(3y\right)}^4+4.{\left(3y\right)}^3.\left(-4\right)+6.{\left(3y\right)}^2.{\left(-4\right)}^2+4.{\left(3y\right)}^1{\left(-4\right)}^3+{\left(-4\right)}^4$

$=81y^4-432y^3+864y^2-768y+256$.

c) ${\left(x+\frac{1}{2}\right)}^4$

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

$=x^4+2x^3+\frac{3}{2}{x}^2+\frac{1}{2}x+\frac{1}{16}$.

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

$=x^4+4.x^3.{\left(-\frac{1}{3}\right)}^1+6.x^2.{\left(-\frac{1}{3}\right)}^2+4.x.{\left(-\frac{1}{3}\right)}^3+{\left(-\frac{1}{3}\right)}^4$

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