\(x^4-3x^3-x+3\)

\(=x^4-x^3-2x^3+2x-3x+3\)

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

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

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

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

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

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

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

\(x^4-3x^3-x+3\)

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

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

x⁴ - 3x³ - x + 3

= (x⁴ - 3x³) - (x + 3)

= x³ (x - 3) - (x - 3)

= (x - 3)(x³ - 1)

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\(x^3-3x^2+6x-4\)

\(=x^3-2x^2+4x-x^2+2x-4\)

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

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

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

x^3 - 3x^2 + 6x - 4 

<=> x^3-3x^2+3x-1+3x-3

<=>(x-1)^3+3(x-1)

<=>(x-1)+((x-1)^2+3)

<=>(x-1)+(x^2-2x+4)

`a)3x-3a+yx-ya`

`=3(x-a)+y(x-a)`

`=(x-a)(y+3)`

`b)x^2-9-4(x+3)`

`=(x-3)(x+3)-4(x+3)`

`=(x+3)(x-3-4)`

`=(x+3)(x-7)`

a)  \(=3x+yx-\left(3a+ya\right)\) \(=x\left(3+y\right)-a\left(3+y\right)\) \(=\left(3+y\right)\left(x-a\right)\)

b) \(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)  \(=\left(x+3\right)\left(x-3-4\right)\) \(=\left(x+3\right)\left(x-7\right)\)

a: \(=3\left(x^2-y^2-x+y\right)\)

\(=3\left[\left(x-y\right)\left(x+y\right)-\left(x-y\right)\right]\)

=3(x-y)(x+y-1)

b: =(x-4)(x+1)

c: =x(x-1)

\(a,=x\left(x-3\right)\\ b,=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\\ c,=\left(3x+7\right)\left(x-2\right)+2\left(x-2\right)=\left(x-2\right)\left(3x+9\right)=3\left(x+3\right)\left(x-2\right)\)

x⁸+3x⁴+4

= x⁸+4x⁴+4-x⁴

= (x⁴+2)²-(x²)²

= (x⁴+2-x²).(x⁴+2+x²)

\(x^8+3x^4+4=(x^8+3x^4)+4=x^4(x^2+3)+4=(x^2+3)(x^4+4)\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

(x-1)(x+2)²