Question

Chứng minh rằng giá trị của biểu thức sau đây bằng 1

A = \(\dfrac{x}{x-3}\) - \(\dfrac{x^2+3x}{2x+3}\). (\(\dfrac{x+3}{x^2-3x}\) - \(\dfrac{x}{x^2-9}\)); với x khác 0, x khác 3; x khác \(\dfrac{3}{2}\)

Answer 1

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

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

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

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

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

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

\(=\dfrac{x-3}{x-3}=1\) ( đpcm )