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!!!
