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 07/12/2021 Bởi Kinsley 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
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!!! Bình luận
$a, (x-1)^2 + (x+3)^2 = 2(x-2)(x+1) + 38$ $⇔ x^2 – 2x +1 + x^2 + 6x +9 = 2x^2 +2x -4x -4 +38$ $⇔ x^2 -2x +x^2 +6x -2x^2 -2x +4x= -4 +38 -10$ $⇔ 6x= 24$ $⇔ x = 4$ Vậy $S=\{4\}$ Bình luận
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!!!
$a, (x-1)^2 + (x+3)^2 = 2(x-2)(x+1) + 38$
$⇔ x^2 – 2x +1 + x^2 + 6x +9 = 2x^2 +2x -4x -4 +38$
$⇔ x^2 -2x +x^2 +6x -2x^2 -2x +4x= -4 +38 -10$
$⇔ 6x= 24$
$⇔ x = 4$
Vậy $S=\{4\}$