a) ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ne0\\9-x^2\ne0\\x^2+2x+1\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ne-1\\x\ne\pm3\end{matrix}\right.\)
b) Ta có:
A = \(\left(1-\frac{4}{x+1}\right):\left(\frac{9-x^2}{x^2+2x+1}\right)\)
A = \(\frac{x+1-4}{x+1}\cdot\frac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x+1}{x+3}\)
a)đkxđ x≠1,-1
bRút gọn
A=\(\left(1-\frac{4}{x+1}\right):\left(\frac{9-x^2}{x^2+2x+1}\right)\)
A=(\(\frac{x+1-4}{x+1}\)).\(\frac{\left(x+1\right)^2}{\left(3-x\right)\left(3+x\right)}\)
A=\(\left(\frac{x-3}{x+2}\right).\)\(\frac{\left(x+1\right)^2}{-\left(x-3\right)\left(x+3\right)}\)
Đến đây tự tính nhé
