Giải bài 4 trang 130 – SGK Toán lớp 8 tập 2

Gợi ý:

$x^4-81=(x^2)^2-9^2$

Đặt $A = \left[\dfrac{x + 3}{(x - 3)^2} + \dfrac{6}{x^2 - 9} - \dfrac{x - 3}{(x + 3)^2}\right].\left[1:\left(\dfrac{24x^2}{x^4 - 81} - \dfrac{12}{x^2 + 9}\right)\right]$
$\begin{align*}A &= \left[\dfrac{x + 3}{(x - 3)^2} + \dfrac{6}{x^2 - 9} - \dfrac{x - 3}{(x + 3)^2}\right].\left[1:\left(\dfrac{24x^2}{x^4 - 81} - \dfrac{12}{x^2 + 9}\right)\right]\\&= \left[\dfrac{(x + 3)^3}{(x - 3)^2(x + 3)^2} + \dfrac{6(x+ 3)(x - 3)}{(x - 3)^2(x + 3)^2} - \dfrac{(x - 3)^3}{(x - 3)^2(x + 3)^2}\right].\left[1:\left(\dfrac{24x^2}{x^4 - 81} - \dfrac{12(x^2 - 9)}{x^4 - 81}\right)\right] \\&= \left[\dfrac{(x + 3)^3 + 6(x - 3)(x + 3) - (x - 3)^3}{(x - 3)^2(x + 3)^2} \right].\left[1:\dfrac{24x^2 - 12(x^2 - 9)}{x^4 - 81} \right] \\ &= \left[\dfrac{x^3 + 9x^2 + 27x + 27 + 6(x^2 - 9) - (x^3 - 9x^2 + 27x - 27)}{(x - 3)^2(x + 3)^2} \right].\left[1:\dfrac{24x^2 - 12x^2 + 108)}{x^4 - 81} \right] \\ &= \left[\dfrac{x^3 + 9x^2 + 27x + 27 + 6x^2 - 54 - x^3 + 9x^2 - 27x + 27)}{(x - 3)^2(x + 3)^2} \right].\left[1:\dfrac{ 12x^2 + 108}{x^4 - 81} \right] \\&= \left[\dfrac{24x^2}{(x - 3)^2(x + 3)^2} \right].\left[\dfrac{ x^4 - 81}{12x^2 + 108}\right] \\&=\left[\dfrac{24x^2}{(x - 3)^2(x + 3)^2} \right].\left[\dfrac{ (x^2 - 9)(x^2 + 9)}{12x^2 + 108}\right] \\&=\dfrac{24x^2}{(x - 3)^2(x + 3)^2} .\dfrac{ (x - 3)(x + 3)(x^2 + 9)}{12(x^2 + 9)} \\& = \dfrac{2x^2}{(x - 3)(x + 3)} \\&= \dfrac{2x^2}{x^2 - 9} \end{align*}$

Với $x = -\dfrac{1}{3},$ ta có: 
$A = \dfrac{2.\left(-\dfrac{1}{3}\right)^2}{\left(-\dfrac{1}{3}\right)^2 - 9} = \dfrac{2.\dfrac{1}{9}}{\dfrac{1}{9} - 9} = -\dfrac{1}{40}$