a, \(\left(1-\frac{x^2-3x}{x^2-9}\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)\) (ĐKXĐ: x \(\ne\pm\) 3; x \(\ne\) 2)
= \(\left(\frac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{\left(3-x\right)\left(3+x\right)}{\left(x+3\right)\left(x-2\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\right)\)
= \(\left(\frac{\left(x-3\right)\left(x+3-x\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{2-x}{x+3}\right)\)
= \(\frac{3}{x+3}:\frac{2-x}{x+3}\)
= \(\frac{3\left(x+3\right)}{\left(x+3\right)\left(2-x\right)}\)
= \(\frac{3}{2-x}\)
Phần b chịu. ngồi nát óc ko ra, không biết đề có lỗi ko chứ mình chịu rồi!
Chúc bn học tốt!!
