Question

Rút gọn biểu thức sau

\(A.\dfrac{x^2-x}{x^2-3x+2}\) \(C.\left(\dfrac{x}{x-2}-\dfrac{2}{x+2}\right):\dfrac{x^2+4}{x+2}\)\(D.\dfrac{x^2-x}{x^2-3x}-\dfrac{7x-9}{x^2-9}\)

Answer 1

\(A.\dfrac{x^2-x}{x^2-3x+2}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{x}{x-2}.\)

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

\(=\dfrac{\dfrac{x^2+4}{x-2}}{x^2+4}=\dfrac{x^2+4}{\left(x-2\right)\left(x^2+4\right)}=\dfrac{1}{x-2}.\)

\(D.\dfrac{x^2-x}{x^2-3x}-\dfrac{7x-9}{x^2-9}=\dfrac{x-1}{x-3}-\dfrac{7x-9}{x^2-9}\)

\(=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{7x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-5x+6}{\left(x-3\right)\left(x+3\right)}\)

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