a,(x^2-2x+1)-4=0 b,x^2-x=-2x+2 c,4x^2+4x+1=x^2 d,x^2-5x+6=0

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1. $#DYHUN$
   `a. (x²-2x+1)-4=0`

   `⇔(x-1)²-2²=0`

   `⇔(x-1-2)(x-1+2)=0`

   `⇔(x-3)(x+1)=0`

   `1)x-3=0⇔x=3`

   `2)x+1=0⇔x=-1`

   Vậy………..

   `b.x²-x=-2x+2`

   `⇔x²-x+2x-2=0`

   `⇔x(x-1)+2(x-1)=0`

   `⇔(x-1)(x+2)=0`

   `1)x-1=0⇔x=1`

   `2)x+2=0⇔x=-2`

   Vậy……..

   `c.4x²+4x+1=x²`

   `⇔4x²+4x+1-x²=0`

   `⇔4x(x+1)-(x²-1)=0`

   `⇔(x+1)(4x-x+1)=0`

   `⇔(x+1)(3x+1)=0`

   `1)x+1=0⇔x=-1`

   `2)3x+1=0⇔x=-1/3`

   Vậy……

   `d.x²-5x+6=0`

   `⇔x²-2x-3x+6=0`

   `⇔x(x-2)-3(x-2)=0`

   `⇔(x-2)(x-3)=0`

   `1)x-2=0⇔x=2`

   `2)x-3=0⇔x=3`

   Vậy…..

    

   Trả lời

2. Đáp án:

   Giải thích các bước giải:

   `a)(x^2-2x+1)-4=0`

   `->(x-1)^2=4`

   `->(x-1)^2=2^2`

   `->` \(\left[ \begin{array}{l}x-1=2\\x-1=-2\end{array} \right.\) 

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

   Vậy `x∈{-1;3}`

   `b)x^2-x=-2x+2`

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

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

   `->(x-1)(x+2)=0`

   `->` \(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\) 

   Vậy `x∈{1;-2}`

   `c)4x^2+4x+1=x^2`

   `->4x^2-x^2+4x+1=0`

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

   `->(3x+1)(x+1)=0`

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

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

   Vậy `x∈{-1/3;-1}`

   `d)x^2-5x+6=0`

   `->(x-2)(x-3)=0`

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

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

   Vậy `x∈{2;3}`

   Trả lời