Question

Khai triển các biểu thức sau

a) (x + 3)²                  b) (3x - 1)²  

c) (x + \(\dfrac{1}{2}\)) (\(\dfrac{1}{2}\) - x)       d) (x² - \(\dfrac{1}{3}\))²

Answer 1

\(a.\left(x+3\right)^2\\ =x^2+2\cdot x\cdot3+3^2\\ =x^2+6x+9\\ b.\left(3x-1\right)^2\\ =\left(3x\right)^2-2\cdot3x\cdot1+1^2\\ =9x^2-6x+1\\ c.\left(x+\dfrac{1}{2}\right)\left(\dfrac{1}{2}-x\right)\\ =\left(\dfrac{1}{2}+x\right)\left(\dfrac{1}{2}-x\right)\\ =\left(\dfrac{1}{2}\right)^2-x^2\\ =\dfrac{1}{4}-x^2\\ d.\left(x^2-\dfrac{1}{3}\right)^2\\ =\left(x^2\right)^2-2\cdot x^2\cdot\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\\ =x^4-\dfrac{2}{3}x^2+\dfrac{1}{9}\)

Answer 2

a) \(\left(x+3\right)^2=x^2+6x+9\)

b) \(\left(3x-1\right)^2=9x^2-6x+1\)

c) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{1}{2}-x\right)=\left(\dfrac{1}{2}+x\right)\left(\dfrac{1}{2}-x\right)=\dfrac{1}{4}-x^2\)

d) \(\left(x^2-\dfrac{1}{3}\right)^2=x^4-\dfrac{2}{3}x^2+\dfrac{1}{9}\)