Question

bài 2 viết biểu thức sau về dạng tích

27x^3-27x^2+3x+1

các bạn giúp mình nhé. thank you

 

Answer 1

Bài 1:

a,27x3+27x2+9x+1a,27x3+27x2+9x+1

=(3x)3+3.(3x)2.1+3.3x.12+13=(3x)3+3.(3x)2.1+3.3x.12+13

=(3x+1)3=(3x+1)3

b,x3+3√2x2y+6xy2+2√2y3b,x3+32x2y+6xy2+22y3

=x3+3.x2.√2y+3.x.(√2y)2+(√2y)3=x3+3.x2.2y+3.x.(2y)2+(2y)3

=(x+√2y)3=(x+2y)3

Bài 2:

a,x3+9x2+27x+27=0a,x3+9x2+27x+27=0

⇔(x+3)3=0⇔(x+3)3=0

⇔x+3=0⇔x=−3⇔x+3=0⇔x=−3

b,(x+1)3−x(x−2)2+x−1=0b,(x+1)3−x(x−2)2+x−1=0

⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0

⇔−x2+8x=0⇔−x2+8x=0

⇔−x(x−8)=0

Answer 2

27x³ - 27x² + 3x + 1

= 27x³ - 9x² - 18x² - 3x + 6x + 1

= (27x³ - 9x²) - (18x² - 6x) - (3x - 1)

= 9x² (3x - 1) - 6x (3x - 1) - (3x - 1)

= (3x - 1) (9x² - 6x - 1)