$(\dfrac{3x+1}{3x-1}-\dfrac{1-3x}{3x+1}).\dfrac{12x}{9x-3}$ ĐK: $x\neq\pm\dfrac{1}{3}$
$=[\dfrac{(3x+1)^2}{(3x-1)(3x+1)}-\dfrac{(1-3x)(3x-1)}{(3x-1)(3x+1)}].\dfrac{12x}{9x-3}$
$=[\dfrac{9x^2+6x+1}{(3x-1)(3x+1)}-\dfrac{-(3x-1)(3x-1)}{(3x-1)(3x+1)}].\dfrac{12x}{9x-3}$
$=\dfrac{9x^2+6x+1+9x^2-6x+1}{(3x-1)(3x+1)}.\dfrac{12x}{3(3x-1)}$
$=\dfrac{12x(18x^2+2)}{3(3x+1)(3x-1)^2}$
$=\dfrac{4x(18x^2+2)}{(3x+1)(3x-1)^2}$
$=\dfrac{8x(9x^2+1)}{(3x+1)(3x-1)^2}$
Thay $x=\dfrac{2006}{3}$ (tmđk) vào biểu thức ta được:
$\dfrac{8.\dfrac{2006}{3}(9.(\dfrac{2006}{3})^2+1)}{(3.\dfrac{2006}{3}+1)(3.\dfrac{2006}{3}-1)^2}$
$=\dfrac{\dfrac{16048}{3}(9.\dfrac{4024036}{9}+1)}{(2006+1)(2006-1)^2}$
$=\dfrac{\dfrac{16048}{3}.4024037}{2007.4020025}$
Bấm máy tính $=2,667998001$.
