## Tìm x 1) (x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x+4)(x-4)+3x² 2) (2x+3)²+(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)² 3) (-x+5)(x-2)+(x-7)(x+7)=(3x+1)²-(3x-2)(3x+

Question

Tìm x
1) (x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x+4)(x-4)+3x²
2) (2x+3)²+(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)²
3) (-x+5)(x-2)+(x-7)(x+7)=(3x+1)²-(3x-2)(3x+2)
4) (5x-1)(x+1)-2(x-3)²=(x+2)(3x-1)-(x+4)²+(x²-x)

in progress 0
8 phút 2021-09-15T22:51:16+00:00 2 Answers 0 views 0

1. Đáp án:

$1) (x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x+4)(x-4)+3x²$

$⇔ x²+5x-5x-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-4x+4x-16)+3x²$

$⇔ x²-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+4x-4x+16+3x²$

$⇔ x²-25-x²-6x-9+3x²-12x+12 – x²-2x-1+x²-4x+4x-16-3x²=0$

$⇔ (x²-x²+3x²-x²+x²-3x²)+(-6x-12x-2x-4x+4x)+(-25-9+12-1-16)=0$

$⇔ -20x-37=0$

$⇔ -20x=39$

$⇔ x=-39/20$

$2) (2x+3)²+(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)²$

$⇔ 4x²+12x+9+x²+x-x-1=5(x²+4x+4)-(x²+x-5x-5)+x²+8x+16$

$⇔ 4x²+12x+9+x²+x-x-1=5x²+20x+20-x²-x+5x+5+x²+8x+16$

$⇔ 4x²+12x+9+x²+x-x-1-5x²-20x-20+x²+x-5x-5-x²-8x-16=0$

$⇔ (4x²+x²-5x²+x²-x²)+(12x+x-x-20x+x-5x-8x)+(9-1-20-5-16)=0$

$⇔ -20x-33=0$

$⇔ -20x=33$

$⇔ x=-1,65$

$3) (-x+5)(x-2)+(x-7)(x+7)=(3x+1)²-(3x-2)(3x+2)$

$⇔ -x²+2x+5x-10+x²+7x-7x-49=9x²+6x+1-(9x²+6x-6x-4)$

$⇔ -x²+2x+5x-10+x²+7x-7x-49 -9x²-6x-1+9x²+6x-6x-4=0$

$⇔ (-x²+x²-9x²+9x²)+(2x+5x+7x-7x-6x+6x-6x)+(-10-49-1-4)=0$

$⇔x-64=0$

$⇔ x=64$

$4) (5x-1)(x+1)-2(x-3)²=(x+2)(3x-1)-(x+4)²+(x²-x)$

$⇔ 5x²+5x-x-1-2(x²-6x+9)=3x²-x+6x-2-(x²+8x+16)+x²-x$

$⇔ 5x²+5x-x-1-2x²+12x-18=3x²-x+6x-2-x²-8x-16+x²-x$

$⇔ 5x²+5x-x-1-2x²+12x-18 – 3x²+x-6x+2+x²+8x+16-x²+x=0$

$⇔ (5x²-2x²-3x²+x²-x²)+(5x-x+12x+x-6x+8x+x)+(-1-18+2+16)=0$

$⇔ 20x-1=0$

$⇔ 20x=1$

$⇔ x=1/20$

BẠN THAM KHẢO NHA!!!

2. Đáp án:

Giải thích các bước giải: