a/ \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\dfrac{2x^3-12x^2+18x+5x^2-30x+45}{3x^3-18x^2+27x-x^2+6x-9}\)
\(=\dfrac{2x\left(x^2-6x+9\right)+5\left(x^2-6x+9\right)}{3x\left(x^2-6x+9\right)-\left(x^2-6x+9\right)}=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)
\(=\dfrac{2x+5}{3x-1}\)
b/ \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\dfrac{x^3+3x^2+2x-2x^2-6x-4}{x^3+3x^2+2x+5x^2+15x+10}\)
\(=\dfrac{x\left(x^2+3x+2\right)-2\left(x^2+3x+2\right)}{x\left(x^2+3x+2\right)+5\left(x^2+3x+2\right)}=\dfrac{\left(x-2\right)\left(x^2+3x+2\right)}{\left(x+5\right)\left(x^2+3x+2\right)}\)
\(=\dfrac{x-2}{x+5}\)
Lời giải:
ĐKXĐ:.........
a) \(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-(x^2-3x)-(15x-45)}{3x^3-9x^2-(10x^2-30x)+(3x-9)}\)
\(=\frac{2x^2(x-3)-x(x-3)-15(x-3)}{3x^2(x-3)-10x(x-3)+3(x-3)}=\frac{(x-3)(2x^2-x-15)}{(x-3)(3x^2-10x+3)}\)
\(=\frac{(x-3)[2x(x-3)+5(x-3)]}{(x-3)[3x(x-3)-(x-3)]}=\frac{(x-3)(x-3)(2x+5)}{(x-3)(x-3)(3x-1)}=\frac{2x+5}{3x-1}\)
b)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2(x+1)-4(x+1)}{x^3+x^2+7x^2+7x+10x+10}\)
\(=\frac{(x+1)(x^2-4)}{x^2(x+1)+7x(x+1)+10(x+1)}=\frac{(x+1)(x-2)(x+2)}{(x+1)(x^2+7x+10)}\)
\(=\frac{(x-2)(x+2)}{x^2+7x+10}=\frac{(x-2)(x+2)}{x(x+2)+5(x+2)}=\frac{(x-2)(x+2)}{(x+2)(x+5)}=\frac{x-2}{x+5}\)