(x-2)^2-x(x+2)=20 5x(x-7)-35+5x=0 nãy tui làm mất ời…

2 bình luận về “(x-2)^2-x(x+2)=20 5x(x-7)-35+5x=0 nãy tui làm mất ời…”

1. a) $(x-2)^{2}-x(x+2)=20$

   <=> $x^{2}-4x+4-x^{2}-2x=20$

   <=> 4-6x = 20

   <=> 6x = 16

   <=> x = $\frac{-8}{3}$ 

   b) 5x(x-7)-35+5x = 0

   <=> 5x(x-7)+5(x-7) = 0

   <=> 5(x+1)(x-7)=0

   <=> \(\left[ \begin{array}{l}x+1=0\\x-7=0\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x=-1\\x=7\end{array} \right.\) 

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2. ????????????????????????????????????????????????????

   $\bullet$  (x-2)^2-x(x+2)=20

   <=> x^2-4x+4-x^2-2x=20

   <=> (x^2-x^2)+(-4x-2x)+4=20

   <=> -6x+4=20

   <=> -6x=16

   <=> x=-8/3

   Vậy x=-8/3

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   $\bullet$  5x(x-7)-35+5x=0

   <=> 5x(x-7)+5x-35=0

   <=> 5x(x-7)+5(x-7)=0

   <=> (x-7)(5x+5)=0

   <=> \(\left[ \begin{array}{l}x-7=0\\5x+5=0\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\) 

   Vậy x in {7;-1}

    

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