a) \(ĐKXĐ:x^2-3x\ne0\)\(\Leftrightarrow x\left(x-3\right)\ne0\)\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne3\end{cases}}\)

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

\(=\frac{\left(x^2-1\right)\left(x-3\right)}{x\left(x-3\right)}=\frac{\left(x-1\right)\left(x+1\right)}{x}\)

b) Với \(x=2\)( thoả mãn ĐKXĐ ) \(\Rightarrow A=\frac{\left(2-1\right)\left(2+1\right)}{2}=\frac{3}{2}\)

1. \(x^2-8x+16=\left(x-4\right)^2\)

2. \(\left(x+5\right)\left(x-5\right)=x^2-25\)

3. \(x^3-6x^2+12x-8\)

\(=\left(x^3-8\right)-\left(6x^2-12x\right)\)

\(=\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)\)

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

\(=\left(x-2\right)^3\)

4. \(=\left(x+2\right)\left(x^2-2x+4\right)=x^3+8\)

5. \(x^2+2x+1=\left(x+1\right)^2\)

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

\(x^2+4x+4\)

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

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

\(=x^3-27\)

a,\(14x^2y-21xy=7xy.\left(2x-3\right)\)

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

c,\(3x^2-6xy+3y^2-27=3.\left(x^2-2xy+y^2-9\right)=3.\text{[}\left(x-y\right)^2-3^2\text{]}=3.\left(x-y-3\right).\left(x-y+3\right)\)

a) 14x²y - 21xy

= 7xy (2x - 3)

b) -x² + 6x - 9

= -(x² - 6x + 3²)

= -(x - 3)²

c) 3x² - 6xy + 3y² - 27

= 3 (x² - 2xy + y² - 9)

= 3 [(x² - 2xy + y²) - 9]

= 3 [(x - y)² - 3²]

= 3 (x - y - 3) (x - y + 3)

#Học tốt!!!

~NTTH~

(m2 - 4)-m +2 =0 ko trắc đúng đâu nha !