\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)

\(x^3-6x^2-24x+144=\left(x^3-6x^2\right)-\left(24x-144\right)=x^2\left(x-6\right)-24\left(x-6\right)=\left(x-6\right)\left(x^2-24\right)\)

\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)

a) 2x²+6x=2x(x+3)
b) x⁴+3x³+x+3=(x⁴+x)+(3x³+3)=x(x³+1)+3(x³+1)=(x+3)(x³+1)
c) 64-x²-y²+2xy=-(x²-2xy+y²)+8²=8-(x+y)²=(8+x+y)(8-x-y)

A= (x+5)(x+1)+(x-2)(x²+2xx+4)-(x²+x-2)
A= x²+6x+5+x³-8-x²-x+2
A= x³+(x²-x²)+(6x-x)+(5-8+2)
A= x³+5x-1

Ta có: \(6x^2-7x+2=6x^2-3x-4x+2\)

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

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

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

\(64-x^2-y^2+2xy=64-\left(x^2-2xy+y^2\right)\)

\(=8^2-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\)

−(x−y−8)(x−y+8)

a: \(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

hay \(x\in\left\{0;\sqrt{3};-\sqrt{3}\right\}\)

b: \(=\dfrac{x^3-3x^2+6x-8}{x-2}=\dfrac{x^2-2x-x^2+2x+4x-8}{x-2}=x^2-x+4\)

a)lát đề

b)x³+x+2

=x³-x²+2x+x²-x+2

=x(x²-x+2)+(x²-x+2)

=(x+1)(x²-x+2)

c)x⁴+64

=(x²)²+8²+2x²*8-2x²*8 

=(x²+8)²-(4x)²

=(x²-4x+8)(x²+4x+8) . 

cái cuối dấu cộng mới biết làm,,