1,Giải Pt a,(x-1) ²+(x+3) ²=2(x-2)(x+1)+38 b,5(x ²-2x-1)+2(3x-2)=5(x+1) ²

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1. Đáp án:

   $a,(x-1) ²+(x+3) ²=2(x-2)(x+1)+38$

   $=> x²-2x+1+x²+6x+9=(2x-4)(x+1)+38$

   $=> x²-2x+1+x²+6x+9=2x²+2x-4x-4+38$

   $=> x²+x²-2x²-2x+6x-2x+4x+1+9+4-38=0$

   $=> 6x-24=0$

   $=> 6x=24$

   $=> x=4$

   Vậy $S=${$4$}

   $b,5(x ²-2x-1)+2(3x-2)=5(x+1) ²$

   $=> 5x²-10x-5+6x-4=5(x²+2x+1)$

   $=> 5x²-10x-5+6x-4=5x²+10x+5$

   $=> 5x²-5x²-10x+6x-10x-5-4-5=0$

   $=> -14x – 14 =0$

   $=> -14x = 14$

   $=> x=-1$

   Vậy $S=${$-1$}

   BẠN THAM KHẢO NHA!!!

    

   Trả lời

2. $a, (x-1)^2 + (x+3)^2 = 2(x-2)(x+1) + 38$

   $⇔ x^2 – 2x +1 + x^2 + 6x +9 = 2x^2 +2x -4x -4 +38$

   $⇔ x^2 -2x +x^2 +6x -2x^2 -2x +4x= -4 +38 -10$

   $⇔ 6x= 24$

   $⇔ x = 4$

   Vậy $S=\{4\}$

   Trả lời