Tìm x biết: `x^3+27-(x+3)(x^2-7x)=0` Giúp eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

2 bình luận về “Tìm x biết: `x^3+27-(x+3)(x^2-7x)=0` Giúp eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee”

1. x^3+27-(x+3)(x^2-7x)=0

   =>(x+3)(x^2-3x+9)-(x+3)(x^2-7x)=0

   =>(x+3)(x^2-3x+9-x^2+7x)=0

   =>(x+3)[(x^2-x^2)+(-3x+7x)+9]=0

   =>(x+3)(4x+9)=0

   =>x+3=0

   4x+9=0

   =>x=0-3

   4x=0-9

   =>x=-3

   4x=-9-> x=-9:4-> x=-9/4

   Vậy x∈{-3;-9/4}

   *HĐT 6: A^3+B^3=(A+B)(A^2-AB+B^2)

   Trả lời
2. Giải đáp:

    x^{3}+27-(x+3).(x^{2}-7x)=0

   => (x^{3}+3^{3})-(x+3).(x^{2}-7x)=0

   => (x+3).(x^{2}-3x+9)-(x+3).(x^{2}-7x)=0

   => (x+3).(x^{2}-3x+9-x^{2}+7x)=0

   => (x+3).(4x+9)=0

   => x+3=0 hoặc 4x+9=0

   => x=-3 hoặc 4x=-9

   => x=-3 hoặc x=-\frac{9}{4}

   Vậy x\in{-3;-\frac{9}{4}}

    

   Trả lời