Question

Rút gọn phân thức:

\(a,\dfrac{8a^{n+2}+a^{n-1}}{16a^{n+4}+4a^{n+2}+a^n}\)

\(b,\dfrac{\left(n+1\right)!-n!}{\left(n+1\right)!+n!}\)

Answer 1

b) \(\dfrac{\left(n+1\right)!-n!}{\left(n+1\right)!+n!}=\dfrac{n!.\left(n+1\right)-n!}{n!\left(n+1\right)+n!}=\dfrac{n!\left(n+1-1\right)}{n!\left(n+1+1\right)}=\dfrac{n}{n+2}\)

a) \(\dfrac{8a^{n+2}+a^{n-1}}{16a^{n+4}+4a^{n+2}+a^n}=\dfrac{8a^{n-1+3}+a^{n-1}}{16a^{n-1+5}+4a^{n-1+3}+a^{n-1+1}}\)

\(=\dfrac{8a^{n-1}.a^3+a^{n-1}}{16a^{n-1}a^5+4a^{n-1}a^3+a^{n-1}a}=\dfrac{a^{n-1}\left(8a^3+1\right)}{a^{n-1}\left(16a^5+4a^3+a\right)}\)

\(=\dfrac{8a^3+1}{16a^5+4a^3+a}\)