Question

a,\(\frac{\left(2x^3\right)}{4x^7}\)

b,\(\frac{\left(x-1\right)}{\left(x+1\right)^2}.\frac{x^2+2x+1}{x^2-1}\)

c,\(\frac{x^2-7x+12}{x^2-16}\)

d, \(\frac{x-1}{\sqrt{x+1}}:\left(\sqrt{x-1}\right)\)

Answer 1

a/ \(\frac{2x^3}{4x^7}=\frac{1}{2x^4}\) với ĐKXĐ : \(x\ne0\)

b/ \(\frac{x-1}{\left(x+1\right)^2}.\frac{x^2+2x+1}{x^2-1}=\frac{x-1}{\left(x+1\right)^2}.\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{1}{x+1}\) với ĐKXĐ : \(x\ne\pm1\)

c/ \(\frac{x^2-7x+12}{x^2-16}=\frac{\left(x-4\right)\left(x-3\right)}{\left(x-4\right)\left(x+4\right)}=\frac{x-3}{x+4}\) với ĐKXĐ : \(x\ne\pm4\)

d/ \(\frac{x-1}{\sqrt{x}+1}:\left(\sqrt{x}-1\right)=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}=1\) với ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)