Question

cho phân thức:

A=x^4-5x^2+4/x^4-x^2+4x-4

a) Rút gọn A

b) Tìm x để A nhận giá trị nguyên

Answer 1

\(a,A=\dfrac{x^4-5x^2+4}{x^4-x^2+4x-4}=\dfrac{x^4-x^2-4x^2+4}{x^2\left(x-1\right)\left(x+1\right)+4\left(x-1\right)}\\ A=\dfrac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{\left(x-1\right)\left(x^2+x+4\right)}\\ A=\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2-4\right)}{\left(x-1\right)\left(x^2+x+4\right)}=\dfrac{\left(x+1\right)\left(x^2-4\right)}{x^2+x+4}=\dfrac{x^3+x^2-4x-4}{x^2+x+4}\)

Answer 2

`1/2`

Answer 3

`1/2

Answer 4

1/2`