9:
a: -x^3+3x^2-3x+1
=(-x)^3+3*(-x)^2*1+3*(-x)*1^2+1^3
=(-x+1)^3
b: z^3-z^2+1/3z-1/27
=z^3-3*z^2*1/3+3*z*(1/3)^2-(1/3)^3
=(z-1/3)^3
c: x^6-3x^4y+3x^2y^2-y^3
=(x^2)^3-3*(x^2)^2*y+3*x^2*y^2-y^3
=(x^2-y)^3
d: =(x-y)^3+3*(x-y)^2*1/3+3*(x-y)*(1/3)^2+(1/3)^3
=(x-y+1/3)^3
Ví dụ 9:
a) \(-x^3+3x^2-3x+1\)
\(=-\left(x^3-3x^2+3x-1\right)\)
\(=-\left(x-1\right)^3\)
b) \(x^3-x^2+\dfrac{1}{3}x-\dfrac{1}{27}\)
\(=x^3-3\cdot\dfrac{1}{3}\cdot x^2+3\cdot\left(\dfrac{1}{3}\right)^2\cdot x-\left(\dfrac{1}{3}\right)^3\)
\(=\left(x-\dfrac{1}{3}\right)^3\)
c) \(x^6-3x^4y+3x^2y^2-y^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot y+3\cdot x^2\cdot y^2-y^3\)
\(=\left(x^2-y\right)^3\)
d) \(\left(x-y\right)^3+\left(x-y\right)^2+\dfrac{1}{3}\left(x-y\right)+\dfrac{1}{27}\)
\(=\left(x-y\right)^3+3\cdot\dfrac{1}{3}\cdot\left(x-y\right)^2+3\cdot\left(\dfrac{1}{3}\right)^2\cdot\left(x-y\right)+\left(\dfrac{1}{3}\right)^3\)
\(=\left(x-y+\dfrac{1}{3}\right)^3\)
Vì dụ 8:
a) \(x^2+6x+...=\left(x+...\right)^2\)
\(\Rightarrow x^2+6x+9=\left(x+3\right)^2\)
b) \(4x^2-4x+...=\left(2x-...\right)^2\)
\(\Rightarrow4x^2-4x+1=\left(2x-1\right)^2\)
c) \(9x^2-...+...=\left(3x-2y\right)^2\)
\(\Rightarrow9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
d) \(\left(x-...\right)\left(...+\dfrac{y}{3}\right)=...-\dfrac{y^2}{9}\)
\(\Rightarrow\left(x-\dfrac{y}{3}\right)\left(x+\dfrac{y}{3}\right)=x^2-\dfrac{y^2}{9}\)
