Question

rút gọn biểu thức 
\(\frac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)

Answer 1

\(\frac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)

\(=\frac{\left(x^4+x^3\right)-\left(x+1\right)}{\left(x^4+x^3\right)+\left(x+1\right)+2x^2}\)

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

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

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

\(=\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)+2x^2}\)

Answer 2

Detpacito làm đúng rồi>_<

Answer 3

Despacito làm sai rồi, vì khi mẫu có dấu cộng ko đc rút gọn
Ta có:

\(\frac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}=\frac{x^3\left(x+1\right)-\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\)

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

k cho mk nha :0 !