Question

Hoàn thành các hằng đẳng thức sau

a) \(\left(3x-1\right)^3\)

b)\(\left(4x-\dfrac{1}{2}\right)^2\)

c)\(\left(\dfrac{1}{3}x+1\right)^3\)

d)\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)^2\)

e) \(x^6-1\)

f) \(27x^3+8\)

g) \(9x^2-4\)

Answer 1

Cái này áp dụng hằng đẳng thức 100%

a, \(\left(3x-1\right)^3=27x^3-3.9x^2+3.3x-1=27x^3-27x^2+9x-1\)

b, \(\left(4x-\dfrac{1}{2}\right)^2=16x^2-2.4x.\dfrac{1}{2}+\dfrac{1}{4}=16x^2-4x+\dfrac{1}{4}\)

c, \(\left(\dfrac{1}{3}x+1\right)^3=\dfrac{1}{27}x^3+3.\dfrac{1}{9}x^2+3.\dfrac{1}{3}x+1=\dfrac{1}{27}x^3+\dfrac{1}{3}x^2+x+1\)

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

e, \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\left(x^3+1\right)\)

f, \(27x^3+8=\left(3x\right)^3+2^3=\left(3x+2\right)\left(9x^2-6x+4\right)\)

g, \(9x^2-4=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\)

Answer 2

a) \(\left(3x-1\right)^3=21x^3-27x^2+9x-1\)

b) \(\left(4x-\dfrac{1}{2}\right)^2=16x^2-4x+\dfrac{1}{4}\)

c) \(\left(\dfrac{1}{3}x+1\right)^3=\dfrac{1}{27}x^3+\dfrac{1}{3}x^2+x+1\)

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

e) \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\times\left(x^3+1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)

f) \(27x^3+8=\left(3x+2\right)\left(9x^2-6x+4\right)\)

g) \(9x^2-4=\left(3x-2\right)\left(3x+2\right)\)