a, \(\dfrac{x^4-4x^2+3}{x^4+6x^2-7}\)=\(\dfrac{x^4-4x^2+4-1}{x^4+7x^2-x^2-7}\)=\(\dfrac{\left(x^2-2\right)^2-1}{x^2\left(x^2+7\right)-\left(x^2+7\right)}\)
=\(\dfrac{\left(x^2-1\right)\left(x^2-3\right)}{\left(x^2-1\right)\left(x^2+7\right)}\)=\(\dfrac{x^2-3}{x^2+7}\)
b, \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)=\(\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{\left(x^2+1\right)^2+x\left(x^2+1\right)}\)=\(\dfrac{\left(x+1\right)\left(x^3-1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}\)=\(\dfrac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}\)=\(\dfrac{x^2-1}{x^2+1}\)
c, \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)=\(\dfrac{x^3+3x^2-3-1}{x^3-3x-1+3}\)=\(\dfrac{\left(x-1\right)\left(x^2+x+1\right)+3\left(x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)-3\left(x-1\right)}\)=\(\dfrac{\left(x-1\right)\left(x^2+x+1+3x+3\right)}{\left(x-1\right)\left(x^2+x+1-3\right)}\)=\(\dfrac{x^2+4x+4}{x^2+x-2}\)=\(\dfrac{\left(x+2\right)^2}{x^2+x-2}\)
Mong là mk làm đúng. Chúc bn may mắn!
