`(x^3 + 3x^2 -4)/(x^3 – 3x +2)` Rút gọn

2 bình luận về “`(x^3 + 3x^2 -4)/(x^3 – 3x +2)` Rút gọn”

1. (x^3+3x^2-4)/(x^3-3x+2)

   = (x^3-x^2+4x^2-4)/(x^3-x-2x+2)

   = [(x^3-x^2)+(4x^2-4)]/[(x^3-x)-(2x-2)]

   = [x^2(x-1)+4(x^2-1)]/[x(x^2-1)-2(x-1)]

   = [x^2(x-1)+4(x+1)(x-1)]/[x(x+1)(x-1)-2(x-1)]

   = [x^2(x-1)+(4x+4)(x-1)]/[(x^2+x)(x-1)-2(x-1)]

   =[(x^2+4x+4)(x-1)]/[(x^2+x-2)(x-1)]

   = [(x+2)^2]/(x^2+x-2)

   = [(x+2)^2]/(x^2+2x-x-2)

   = [(x+2)^2]/[(x^2+2x)-(x+2)]

   = [(x+2)^2]/[x(x+2)-1.(x+2)]

   = [(x+2)^2]/[(x-1)(x+2)]

   = [(x+2)(x+2)]/[(x-1)(x+2)]

   = (x+2)/(x-1)

    

   Trả lời
2. Giải đáp:

   \frac{x^{3}+3x^{2}-4}{x^{3}-3x+2} (x\ne1)

   =\frac{x^{3}+4x^{2}-x^{2}-4}{x^{3}-2x-x+2}

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

   =\frac{x^{2}.(x-1)+4.(x-1).(x+1)}{x.(x^{2}-1)-2.(x-1)}

   =\frac{x^{2}.(x-1)+(4x+4).(x-1)}{x.(x-1).(x+1)-2.(x-1)}

   =\frac{(x^{2}+4x+4).(x-1)}{(x^{2}+x).(x-1)-2.(x-1)}

   =\frac{(x+2)^{2}.(x-1)}{(x^{2}+x-2).(x-1)}

   =\frac{x^{2}+4x+4}{x^{2}+x-2}

   =\frac{x^{2}+2.x.2+2^{2}}{x^{2}+2x-x-2}

   =\frac{(x+2)^{2}}{x.(x+2)-1.(x+2)}

   =\frac{(x+2)^{2}}{(x-1).(x+2)}

   =\frac{x+2}{x-1}

   #DYNA

    

   Trả lời