câu 1:

x³-1+3x²-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)

Câu 2 :

a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)

\(=\left(x^4-x^2-2x^3+2x+x^2-1\right):\left(x^2-1\right)\)

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

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

\(=x^2-2x+1\)

b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)

\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)

Câu 3 :

Sửa đề :

\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)

ĐKXĐ : \(x\ne2\)

\(\frac{3x^2+6x+12}{x^3-8}\)

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

\(=\frac{3}{x-2}\)

3x²+6x+12/x³-8

=3*(x^2+2x+4)/x^3-2^3

=3*(x^2+2x+4)/(x-2)*(x^2+2x+4)

=3/x-2

a,ĐKXĐ \(x^3-8\ne0\Leftrightarrow x^3\ne8\Leftrightarrow x\ne2\)

b,\(\Leftrightarrow3x^2+6x+12=0\)

    \(\Leftrightarrow3\left(x^2+2x+1\right)+9=0\)

   \(\Leftrightarrow3\left(x+1\right)^2+9=0\)(VÔ LÝ VÌ 3(x+1)²>=0 =>3(x+1)²+9>0)

vì vây ko có giá trị x để F =0

C, VỚI ĐKXĐ trên ,ta có 

\(F=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

    \(=\frac{3}{x-2}\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

a) Để phân thức trên xác định \(\Leftrightarrow x^3-8\ne0\Leftrightarrow x\ne2\)

b) \(\frac{3x^2+6x+12}{x^3-8}\)

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

\(=\frac{3}{x-2}\)

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)

Mik cảm ơn ạ 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)