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

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

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

a, \(\dfrac{27}{8x^3-1}:\dfrac{3}{2x-1}\)

\(=\dfrac{27}{\left(2x-1\right)\left(4x^2+2x+1\right)}.\dfrac{2x-1}{3}\)

\(=\dfrac{9}{4x^2+2x+1}\)

b, \(\dfrac{8x^3+36x^2+54x+27}{2x+3}=\dfrac{\left(2x+3\right)^3}{2x+3}=\left(2x+3\right)^2\)

\(36x^2+36x+9=\left(6x\right)^2+36x+3^2=\left(6x+3\right)^2\)

36x²+36x+9=(6x)²+2.3.6x+3²=(6x+3)²

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

b)\(x^4-2x^3-x^2+2x+1=x^4+\left(-x\right)^2+\left(-1\right)^2+2x^2\left(-x\right)+2.\left(-x\right).\left(-1\right)+2x^2.\left(-1\right)\)

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

c)\(8x^3+27y^3+36x^2y+54xy^2=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)

\(=\left(2x+3y\right)^2\)

\(B=\left(\dfrac{1}{2}x-1\right)^3=\left(-1-1\right)^3=\left(-2\right)^3=-8\)

\(C=\left(x+1\right)^3-1000\)

\(=100^3-1000=999000\)

\(D=27x^3+54x^2+36x+8-4\)

\(=\left(3x+2\right)^3-4=\left(-6+2\right)^3-4\)

\(=-64-4=-68\)

8x³- 125= (2x)³- 5³= (2x-5)[(2x)²+2x5+5² ]=(2x-5)(4x²+10x+25)

\(8x^3-125\)

\(=\left(2x\right)^3-5^3\)

\(=\left(2x-5\right)\left(4x^2+10x+25\right)\)

Áp dụng hằng đẳng thức thứ 7

\(8x^3-125\)
\(=\left(2x\right)^3-5^3\)

\(\left(2x-5\right)\left(4x^2-10x+25\right)\)

\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

\(=\left(2\cdot24.5+1\right)^3=50^3=125000\)

a) 25x² + 10xy + y²

= (5x)² + 2 . 5x . y + y²

= (5x + y)²

b) 8x³ + 36x²y + 54xy² + 27y³

= (2x)³ + 3 . (2x)² . 3y + 3 . 2x . (3y)² + (3y)³

= (2x + 3y)³

1) \(B=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2+\dfrac{3}{2}x-1=\left(\dfrac{1}{2}x-1\right)^3\)

thay x =-2 vào B, ta được:

\(B=\left(\dfrac{1}{2}\cdot\left(-2\right)-1\right)^3=\left(-2\right)^3=-8\)

2) \(C=x^3+3x^2+3x-999=\left(x+1\right)^3-1000\)

thay x =99 vào B, ta được:

\(C=\left(99+1\right)^3-1000=999000\)

3) \(D=27x^3+54x^2+36x+4=\left(3x+2\right)^3-4\)

thay x =-2 vào D, ta được:

\(D=\left(3\left(-2\right)+2\right)^3-4=-68\)