tìm x, biết: a, x(+x1)-x(x-3)=0 b,x ²-6x+8=0

2 bình luận về “tìm x, biết: a, x(+x1)-x(x-3)=0 b,x ²-6x+8=0”

1. a) x(x+1)-x(x-3)=0

   ⇒ x^2+x-x^2+3x=0

   ⇒ x^2-x^2+x+3x=0

   ⇒ 4x=0

   ⇒ x=0

   Vậy x=0

   b) x^2-6x+8=0

   ⇒ x^2-4x-2x+8=0

   ⇒ (x^2-4x)-(2x-8)=0

   ⇒ x(x-4)-2(x-8)=0

   ⇒ (x-4)(x-2)=0

   ⇒ \(\left[ \begin{array}{l}x-4=0\\x-2=0\end{array} \right.\)

   ⇒ \(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\) 

   Vậy x∈{4;2}

    

   Trả lời
2. a) x(x+1)-x(x-3)=0

   => x^2+x-x^2+3x=0

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

   => 4x=0

   => x=0

   Vậy x=0

   b) x^2-6x+8=0

   => x^2-2x-4x+8=0

   => (x^2-2x)-(4x-8)=0

   => x(x-2)-4(x-2)=0

   => (x-4)(x-2)=0

   TH1:

   x-4=0

   => x=4

   TH2:

   x-2=0

   => x=2

   Vậy x∈{4; 2}

    

   Trả lời