1. Ta có: \(\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)
\(=\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)}\)
\(=\dfrac{x^4\left(x^2+1\right)+x^2+1}{x-1}\)
\(=\dfrac{\left(x^2+1\right)\left(x^4+1\right)}{x-1}\)
2.Ta có: \(\dfrac{x^2+y^2+z^2-2xy+2xz-2xz}{x^2-2xy+y^2-z^2}\)
\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\dfrac{\left(x-y+z\right)\left(x-y+z\right)}{\left(x-y-z\right)\left(x-y+z\right)}=\dfrac{x-y+z}{x-y-z}\)
_Chúc bạn học tốt_
\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{\left(x-1\right)}\\ \)
\(\text{2) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2z\left(x-y\right)+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)