a) \(A=\frac{x^3-3x^2+3x-1}{x^2-1}=\frac{x^3-3x^2.1+3x.1^3-1^3}{x^2-1^2}=\frac{\left(x-1\right)^3}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x-1\right)^2}{x+1}\)
b) TH1 : Thay x=5 vào ta có :
\(A=\frac{\left(x-1\right)^2}{x+1}=\frac{\left(5-1\right)^2}{5+1}=\frac{4^2}{5}=\frac{16}{5}\)
TH2 : Thay x=-5 vào ta có :
\(A=\frac{\left(x-1\right)^2}{x+1}=\frac{\left(-5-1\right)^2}{-5+1}=\frac{36}{-4}=-9\)
a) A=\(\frac{x^3-3x^2+3x-1}{x^2-1}\)
A=\(\frac{x^3-3x^2.1+3x.1^2-1^3}{x^2-1^2}\)
A=\(\frac{\left(x-1\right)^3}{\left(x-1\right)\left(x+1\right)}\)
A=\(\frac{\left(x-1\right)\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
A=\(\frac{\left(x-1\right)^2}{x+1}\)
b) Ta có: \(|x|=5\)
=>x=5 hoặc x=-5
TH1:Nếu x=5 thì :A=\(\frac{\left(5-1\right)^2}{5+1}=\frac{4^2}{6}=\frac{16}{6}=\frac{8}{3}\)
TH2:Nếu x=-5 thì:A=\(\frac{\left(-5-1\right)^2}{-5+1}=\frac{\left(-6\right)^2}{-4}=\frac{36}{-4}=-9\)
