Phân tích đa thức thành nhân tử a) 3x.(x-1)=x-1 b) 2(x+5)-x^2-5x=0

2 bình luận về “Phân tích đa thức thành nhân tử a) 3x.(x-1)=x-1 b) 2(x+5)-x^2-5x=0”

1. a)

   3x(x-1)=x-1

   <=> 3x(x-1)-(x-1)=0

   <=> (x-1)(3x-1)=0

   <=> \(\left[ \begin{array}{l}x-1=0\\3x-1=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=1\\x=\dfrac{1}{3}\end{array} \right.\) 

   Vậy x∈{1;1/3}

   b)

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

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

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

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

   Vậy x∈{-5;2}

   $#<33$

   Trả lời
2. a)

   3x.(x-1)=x-1

   <=>3x.(x-1)-(x-1).1=0

   <=>(3x-1).(x-1)=0

   <=>[(3x-1=0),(x-1=0):}

   <=>[(x=1/3),(x=1):}

   Vậy x in {1/3;1}

   b)

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

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

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

   <=>[(2-x=0),(x+5=0):}

   <=>[(x=2),(x=-5):}

   Vậy x in {2;-5}

   Trả lời