Question

Chứng minh đa thức sau x^4+x^3+x+1/ x^4-x^3+2x^2-x+1 = (x+1)^2 /x^2+1

Answer 1

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

Ta có :

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

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

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

Chúc bạn học tốt !!!