Question

rút gọn biểu thức A=\(\left(\dfrac{10x-5}{x^2-9}-\dfrac{3-x}{x+3}\right):\dfrac{x+2}{x+3}\)

Answer 1

\(A=\left(\dfrac{10x-5}{x^2-9}-\dfrac{3-x}{x+3}\right):\dfrac{x+2}{x+3}\)

= \(\left(\dfrac{10x-5}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-x}{x+3}\right):\dfrac{x+2}{x+3}\)

= \(\left(\dfrac{10x-5}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(3-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right).\dfrac{x+2}{x+3}\)

=\(\dfrac{10x-5+x^2-6x+9}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)

=\(\dfrac{x^2+4x+4}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)

=\(\dfrac{\left(x+2\right)^2}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)

=\(\dfrac{x+2}{x-3}\)

Answer 2

\(A=\left(\dfrac{10x-5}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-3}{x+3}\right).\dfrac{x+3}{x+2}\)

<=>\(\left\{{}\begin{matrix}x\ne-3\\A=\dfrac{10x-5}{\left(x-3\right)\left(x+2\right)}+\dfrac{x-3}{\left(x+3\right)\left(x+2\right)}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x\ne\pm3;-2\\A=\dfrac{\left(10x-5\right)\left(x+3\right)+\left(x-3\right)^2}{\left(x-3\right)\left(x+2\right)\left(x+3\right)}=\dfrac{11x-3}{x^2-9}\end{matrix}\right.\)