Tìm x, biết: x.( x – 4 ) + 1 = 3x – 5

2 bình luận về “Tìm x, biết: x.( x – 4 ) + 1 = 3x – 5”

1. x(x-4)+1=3x-5

   => x^2 – 4x+1-3x+5=0

   => x^2 -7x +6=0

   => x^2 -x -6x +6=0

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

   => x(x-1)-6(x-1)=0

   => (x-6)(x-1)=0

   => $[\begin{array}{c}x-6=0\\ x-1=0\end{array}$

   => $[\begin{array}{c}x=6\\ x=1\end{array}$

   Vậy x in {1;6}

   Trả lời
2. x(x-4)+1=3x-5

   => x^2-4x+1=3x-5

   => x^2-4x+1-3x+5=0

   => x^2-(4x+3x)+(1+5)=0

   => x^2-7x+6=0

   => x^2-x-6x+6=0

   => x(x-1)-6(x-1)=0

   => (x-6)(x-1)=0

   TH1:

   x-6=0

   => x=6

   TH2:

   x-1=0

   => x=1

   Vậy x={6;1}

   Trả lời