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

\(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2-2.2x^4+1+2.2.x^4=\left(2x^4+1\right)^2-4x^4\)

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

\(x^2-8x-9==x^2+x-9x-9=x\left(x+1\right)-9\left(x+1\right)=\left(x+1\right)\left(x-9\right)\)

\(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)

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

b) \(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2-2.2x^4+1+2.2.x^4=\left(2x^4+1\right)^2-4x^4\)

c) \(x^2-8x-9==x^2+x-9x-9=x\left(x+1\right)-9\left(x+1\right)=\left(x+1\right)\left(x-9\right)\)

d) \(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)

a) x² + 7x + 12 = x² + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )

b) x² - 10x + 16 = x² - 2x - 8x + 16 = x( x - 2 ) - 8( x - 2 ) = ( x - 2 )( x - 8 )

c) x² + 6x + 8 = x² + 2x + 4x + 8 = x( x + 2 ) + 4( x + 2 ) = ( x + 2 )( x + 4 )

d) x² - 8x + 15 = x² - 3x - 5x + 15 = x( x - 3 ) - 5( x - 3 ) = ( x - 3 )( x - 5 )

e) x² - 8x - 9 = x² + x - 9x - 9 = x( x + 1 ) - 9( x + 1 ) = ( x + 1 )( x - 9 )

f) x² + 14x + 48 = x² + 6x + 8x + 48 = x( x + 6 ) + 8( x + 6 ) = ( x + 6 )( x + 8 )

c) x² - x + 9x - 9

=(x²-x) +  (9x-9)

= x(x-1) + 9(x-1)

= (x-1)(x+9)

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

b) \(4x^8+1=4x^8+4x^4+1-4x^4=\left(2x^4+1\right)^2-4x^4=\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)

d) \(x^2+14x+48=\left(x+7\right)^2-1=\left(x+7+1\right)\left(x+7-1\right)=\left(x+8\right)\left(x+6\right)\)

\(x^2-x-12\)

\(=x^2+3x-4x-12\)

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

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

\(x^3-5x^2+8x-4\)

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

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

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

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

a) -ĐKXĐ của A:

x+3≠0 ⇔x≠-3.

x²-9≠0 ⇔(x-3)(x+3)≠0 ⇔x-3≠0 hay x+3≠0⇔x≠3 hay x≠-3.

x-3≠0 ⇔x≠3.

b) B=x²+5x+6=x²+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)

c) A=\(\dfrac{x}{x+3}-\dfrac{6x}{x^2-9}+\dfrac{2}{x-3}\)=\(\dfrac{x\left(x-3\right)+2\left(x+3\right)-6x}{\left(x+3\right)\left(x-3\right)}\)=\(\dfrac{x^2-3x+2x+6-6x}{\left(x+3\right)\left(x-3\right)}\)=\(\dfrac{x^2-7x+6}{x^2-9}\)

d)- Vì x=37 thỏa mãn ĐKXĐ của A và A=\(\dfrac{x^2-7x+6}{x^2-9}\)nên:

A=\(\dfrac{37^2-7.37+6}{37^2-9}=\dfrac{279}{340}\)

\(=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+6\right)\left(x+4\right)\left(x+2\right)\left(x+8\right)+16\)

\(=\left(x^2+10x+24\right)\left(x^2+10x+16\right)+16\)

\(=\left(t+8\right)t+16=\left(t+4\right)^2=\left(x^2+10x+20\right)^2\)

\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

\(8x\left(x^2-9\right)=0\Rightarrow8x\left(x-3\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)