a) \(x^3 + 1 = (x + 1)(x^2 - x + 1)\)
    \(x^9 + x^7 - 3x^2 - 3 = x^7(x^2 + 1) - 3(x^2 + 1) = (x^2 + 1)(x^7 - 3)\).
Điều kiện của x để giá trị của biểu thức Q xác định là \(x \neq -1, x^7 \neq 3, x \neq -3, x \neq 4\).

b) \(Q = \left[\frac{x^7 -3}{x^3 + 1}.\frac{(x - 1)(x + 1)(x^2 - x + 1)}{(x^7 - 3)(x^2 + 1)} + 1 - \frac{2(x + 6)}{x^2 + 1}\right].\frac{(2x + 1)^2}{(x + 3)(4 - x)}\)
     \(= \left[\frac{x^7 - 3}{x^3 + 1}.\frac{(x - 1)(x^3 + 1)}{(x^7 - 3)(x^2 + 1)} + 1 - \frac{2(x + 6)}{x^2 + 1}\right].\frac{(2x + 1)^2}{(x + 3)(4 - x)}\)

giải pt , \(\sqrt{x^4+4x^2}+\sqrt{x+x^2}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}.\)

\(x=0\)

\(x^3=0\)

\(x^3=2.0.\sqrt{0}\)

\(x^3=2x\sqrt{x}\)

\(x^3=2x\sqrt{x}\)     

\(4\left(x^3-2x\sqrt{x}\right)^2=0\)

\(4\left(x^6-4x^4\sqrt{x}+4x^2x\right)=0\)

\(4x^6-16x^4\sqrt{x}+16x^2x=0\)

\(4x^6+16x^3=16x^4\sqrt{x}\)

\(16x^4+4x^5+4x^6+16x^3=16x^4+4x^5+16x^4\sqrt{x}\)

\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(4x^4+4x^4\sqrt{x}+x^4.x\right)\)

\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(2x^2+x^2\sqrt{x}\right)^2\)

\(2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)\)

\(x^4+x^2+4x^2+x+2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)+x^4+x^2+4x^2+x\)

\(\left(\sqrt{x^4+4x^2}+\sqrt{x^2+x}\right)^2=\left(x^4+2x^2\sqrt{x}+x\right)+9x^2\)

\(\sqrt{x^4+4x^2}+\sqrt{x^2+x}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}\)

vậy x=0 là nghiệm của pt  =)) 

1.     \(\frac{x^3-10x^2+25x}{x^2-5x}\)\(=0\)  ( đkxđ: \(x\ne0;5\))

<=> \(\frac{x\left(x-5\right)^2}{x\left(x-5\right)}=0\)<=>   \(x-5=0\)<=> vô n_o

_{2.       \(A=\)\(\frac{2x^2-2}{x^3-x^2-4x+4}\)\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)}   ( a, đkxđ: \(x\ne1;\pm2\))

b, \(A=0\)<=> \(\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=0\)<=> \(x=-1\)( TM) . Vậy \(A=0\Leftrightarrow x=-1\)

3.    \(B=\frac{3x^2-12}{\left(x-3\right)\left(x^2+4x+4\right)}\)\(=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)^2}\) ( a,  đkxđ: \(x\ne3,-2\))

\(b,B=0\Leftrightarrow\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}=0\Leftrightarrow x=2\left(tm\right)\). Vậy \(B=0\Leftrightarrow x=2\)