Tim x: a/ (x-3)^2+x(2-x)=1 b/ 3x(x-5)+2x-10=0

2 bình luận về “Tim x: a/ (x-3)^2+x(2-x)=1 b/ 3x(x-5)+2x-10=0”

1. \bb a)

   (x-3)^2+x.(2-x)=1

   <=>x^2-6x+9+2x-x^2=1

   <=>-4x=1-9

   <=>-4x=-8

   <=>x=2

   Vậy x=2

   $\\$

   \bb b)

   3x.(x-5)+2x-10=0

   <=>3x.(x-5)+2.(x-5)=0

   <=>(x-5).(3x+2)=0

   <=>\(\left[ \begin{array}{l}x-5=0\\3x+2=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=5\\x=-\dfrac{2}{3}\end{array} \right.\) 

   Vậy x={5; -2/3}

   Trả lời
2. a/ (x-3)^2+x(2-x)=1

   x^2 – 6x + 9 + 2x – x^2 = 1

   (x^2 – x^2)+(-6x+2x) = 1 – 9

   -4x = -8

   x = $\dfrac{-8}{-4}$

   x = 2

   b/ 3x(x-5)+2x-10=0

   3x(x-5) + 2(x-5) = 0

   (x-5) (3x+2) = 0

   x-5= 0 hoặc 3x + 2 = 0

   x = 5 hoặc 3x = -2

   x = 5 hoặc x = $\dfrac{-2}{3}$

   Trả lời