tìm x: (x-2)^2+5x*(x-2)=0 c/m: (x-y)^3-(x-y)*(x^2+xy+y^2)+3xy*(x-y)=0

2 bình luận về “tìm x: (x-2)^2+5x*(x-2)=0 c/m: (x-y)^3-(x-y)*(x^2+xy+y^2)+3xy*(x-y)=0”

1. 1)

   (x-2)^2+5x(x-2)=0

   <=>(x-2)(x-2+5x)=0

   <=>(x-2)(6x-2)=0

   <=>\(\left[ \begin{array}{l}x-2=0\\6x-2=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{3}\end{array} \right.\) 

   Vậy x in {2;1/3}

   2)

   (x-y)^3-(x-y)(x^2+xy+y^2)+3xy(x-y)=0

   <=>(x-y)^2=(x-y)(x^2+xy+y^2)-3xy(x-y)

   <=>x^3-3x^2y+3xy^3-y^3=x^3-y^3-3x^2y+3xy^2

   <=>x^3-3x^2y+3xy^3-y^3=x^3-3x^2y+3xy^2-y^3(đúng)

   Vậy (x-y)^3-(x-y)(x^2+xy+y^2)+3xy(x-y)=0

   Trả lời
2. Tìm x:

   (x-2)^2+5x(x-2)=0

   <=> x^2-4x+4+5x^2-10x=0

   <=> 6x^2-14x+4=0

   <=> 6x^2-12x-2x+4=0

   <=> (6x^2-12x)-(2x-4)=0

   <=> 6x(x-2)-2(x-2)=0

   <=> (6x-2)(x-2)=0

   TH1:

   6x-2=0

   <=> 6x=2

   <=> x=1/3

   TH2:

   x-2=0

   <=> x=2

   Vậy x∈{1/3; 2}

   c/m:

   -> VT: (x-y)^3-(x-y)(x^2+xy+y^2)+3xy(x-y)

   =(x-y)[(x-y)(x-y)-(x^2+xy+y^2)+3xy]

   =(x-y)(x^2-xy-xy+y^2-x^2-xy-y^2+3xy)

   =(x-y)[(x^2-x^2)-(xy+xy+xy-3xy)+(y^2-y^2)]

   =(x-y)*0

   =0

   Vậy (x-y)^3-(x-y)(x^2+xy+y^2)+3xy(x-y)=0 (đpcm)

    

   Trả lời