\(\frac{x+3}{x-2}+6-\left(\frac{x-3}{x+2}\right)^2-7\left(\frac{x^2-9}{x^2-4}\right)=0\)
điều kiện xác định X khác (-2,-3,2,3)
<=> \(\frac{\left(x+3\right)\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)^2}-\frac{\left(x-3\right)^2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)^2}-\frac{7\left(x^2-3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)^2}=0\)
=> \(\left(x+3\right)\left(x+2\right)^2-\left(x-3\right)^2\left(x-2\right)-7\left(x^2-9\right)\left(x-2\right)=0\)
<=>\(\left(x+3\right)\left(x^2+4x+4\right)-\left(x^2-6x+9\right)\left(x-2\right)-7\left(x^3-2x^2-9x+18\right)=0\)
\(x^3+7x^2+16x+12-x^3+8x^2-21x+18-7x^3+14x^2+63x-126=0\)
<=> \(-7x^3+29x^2+58x-96=0\)
giải pt trên rồi kết họp đk là xong
