Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a/ x² + 4x - 21= x² - 3x +4x - 21

                       = (x²+4x)-(3x+21)

                       = x(x+4)- 3(x+7)

                       = (x-3).(x+7)

 b/ 3x²-6xy+3y²-3z² = 3(x²- 2xy+y²- z²)

                                  = 3[(x² + 2xy + y²) – z²]

                                  = 3[(x + y)² – z²]

                                  = 3(x + y – z)(x + y + z)

  c/ 2x²y + 12xy + 18y = 2y(x²+6x+9)

n³ -n² -7n +1=n²(n-1) - 7(n-1)-6=0 <=>(n-1).(n²-7)=6

a: \(=\left(x+1\right)\left(x^2-x+1\right)+5x\left(x+1\right)\)

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

a: =2(x-2)+y(x-2)

=(x-2)(2+y)

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

c: =(x-7)(x+2)

a.

2x - 4 + xy - 2y

= 2(x-2) +y(x-2)

= (x-2)(y+2)

c.

x^2 - 5x - 14

= x^2 + 2x - 7x - 14

= x(x+2) - 7(x+2)

= (x-7)(x+2)

Câu 1: A

Câu 21: A

\(16,A\\ 17,C\\ 18,A\\ 19,C\\ 20,A\\ 21,A\)

1a

`a^{3}+3a^{2}-6a-8`

`=a^{3}-8+3a(a-2)`

`=(a-2)(a^{2}+2a+4)+3a(a-2)`

`=(a-2)(a^{2}+2a+4+3a)`

`=(a-2)(a^{2}+5a+4)`

`=(a-2)(a+1)(a+4)`

\(a^3-8+3a\left(a-2\right)\)

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

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

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

\(=\left(a-2\right)\left(a^2+5a+4\right)\)

\(\left(a-2\right)\left(a+1\right)\left(a+4\right)\)

\(a^3+3a^2-6a-8\)

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

\(=\left(a-2\right)\left(a^2+5a+4\right)\)

\(=\left(a-2\right)\left(a+1\right)\left(a+4\right)\)