Đáp án:
Giải thích các bước giải:
Đặt `A=[(x+3)/((x-3)^2)+6/(x^2-9)-(x-3)/((x+3)^2)][1:((24x^2)/(x^4-81)-12/(x^2+9))]`
`ĐKXĐ: x \ne 9, x \ne ±3`
`A=[\frac{(x+3).(x+3)^2}{(x-3)^2.(x+3)^2}+\frac{6.(x^2-9)}{(x-3)^2.(x+3)^2}-\frac{(x-3).(x-3)^2}{(x-3)^2.(x+3)^2}]:[1:(\frac{24x^2}{(x^2-9).(x^2+9)}-\frac{12.(x^2-9)}{(x^2-9).(x^2+9)})]`
`A=[\frac{x^3+9x^2+9x+27+6x^2-54-x^3+9x^2-9x+27}{(x-3)^2.(x+3)^2}]:[1:\frac{24x^2-12x^2+108}{(x^2-9).(x^2+9)}]`
`A=[\frac{24x^2}{(x-3)^2.(x+3)^2}]:[1:(\frac{12x^2+108}{(x^2-9).(x^2+9)}]`
`A=[\frac{24x^2}{(x-3)^2.(x+3)^2}]:(1:\frac{12}{x^2-9})`
`A=\frac{24x^2}{(x-3)^2.(x+3)^2}.\frac{x^2-9}{12}`
`A=\frac{2x^2}{x^2-9}`
