bài 2, e.\(x^3-3x^2+3x-1\)
=\(x^3-x^2-2x^2+2x+x-1\)
=\(\left(x^3-x^2\right)\)-\(\left(2x^2-2x\right)\)+(x-1)
=\(x^2\left(x-1\right)\)-2x(x-1)+(x-1)
=(x-1)(x\(^2\)-2x+1)
=(x-1)\(^3\)
h. \(x^3+1-x^2-x\)
=(x\(^3\)-x\(^2\))-(x-1)
=x\(^2\)(x-1)-(x-1)
=(x-1)(x\(^2\)-1)
g. \(x^3+6x^2+12x+8\)
=\(x^3+2x^2+4x^2+8x+4x+8\)
=\(\left(x^3+2x^2\right)+\left(4x^2+8x\right)+\left(4x+8\right)\)
=\(x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)
=(x+2)(\(x^2+4x+4\))
=(x+2)\(^3\)
k.\(\left(x+y\right)^3\) -x\(^3\)-y\(^3\)
= \(\left(x^3+3x^2y+3xy^2+y^3\right)-x^3-y^3\)
=\(x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
=\(3x^2y+3xy^2\)
=3xy(x+y)
bài 3, a. \(4x^2-49=0\)
\(4x^2=49\)
x\(^2\)=\(\frac{49}{4}\)
x=√\(\frac{49}{4}\)
x=\(\frac{7}{2}\)
vậy x=\(\frac{7}{2}\)
