Question

\(A=\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x+3}\)

rút gọn A

Answer 1

ĐKXĐ \(\hept{\begin{cases}x\ne3\\x\ne-3\\x\ne0\end{cases}}\)

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

\(=\frac{x^2+x}{x^2-3x}\)

Answer 2

Ta có:  A = \(\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x+3}\)
\(\Leftrightarrow\) A = \(\left(\frac{x^2-3}{\left(x-3\right)\left(x+3\right)}+\frac{1}{x-3}\right):\frac{x}{x+3}\)
\(\Leftrightarrow\) A = \(\left(\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x}\)
\(\Leftrightarrow\) A = \(\frac{x^2+x}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x}\)
\(\Leftrightarrow\) A = \(\frac{x^2+x}{x^2-3x}\)