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

ra \(\frac{3}{x-2}\) đúng k ạ............

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,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 

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

    \(=\frac{3}{x-2}\)

Trả lời:

a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)

d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)

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

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

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

a) Điều kiện:
          x³ - 8 \(\ne\)0
\(\Leftrightarrow\)(x - 2)(x² + 2x + 4)\(\ne\)0
\(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x^2+2x+4\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne2\\\left(x+1\right)^2+3\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne2\\\left(x+1\right)^2\ne-3\end{cases}}\)
(vô lí vì (x + 1)² \(\ge\)0 > -3)
\(\Rightarrow\)x \(\ne\)2

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

c) Thế x = \(\frac{4001}{2000}\)vào, ta có:
\(\frac{3x^2+6x+12}{x^3-8}\)
\(=\frac{3}{x-2}\)
\(=\frac{3}{\frac{4001}{2000}-2}\)
\(=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}\)
\(=\frac{3}{\frac{1}{2000}}\)
\(=3.2000=6000\)

Bài 1 :

\(2x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=5\end{cases}}\)

KL :...

bài 2 :

\(x^2+6x+9-y^2=\left(x+3\right)^2-y^2\)

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

Bài 3 :

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

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