Đáp án:
Bài 6 :
$1) 4x(x-5) - (x-1)(4x-3)=5$
$⇔4x^2-20x -4x^2+3x+4x-3 =5$
$⇔ 4x^2-4x^2 -20x +3x+4x =5 + 3$
$⇔ -13x = 8$
$⇔ x = -\dfrac{8}{13}$
Vậy $x=-\dfrac{8}{13}$
$2) (3x-4)(x-2) = 3x(x-9) -3$
$⇔ 3x^2-6x-4x+8 = 3x^2-27x-3$
$⇔ 3x^2-3x^2-6x-4x+27x = -3-8$
$⇔17x = -11$
$⇔ x = -\dfrac{11}{17}$
Vậy $x=-\dfrac{11}{17}$
$3) (x-5)(x-4) - (x+1)(x-2) = 7$
$⇔ x^2-4x-5x+20 -x^2+2x-x+2 = 7$
$⇔x^2-x^2-4x-5x+2x-x = 7-2-20$
$⇔-8x = -15$
$⇔ x = \dfrac{15}{8}$
Vậy $x=\dfrac{15}{8}$
$4) 5x(x-3) =(x-2)(5x-1) -5$
$⇔ 5x^2-15x =5x^2-x-10x+2 - 5$
$⇔ 5x^2 -5x^2-15x+x+10x = 2-5$
$⇔-4x = -3$
$⇔ x = \dfrac{3}{4}$
Vậy $x=\dfrac{3}{4}$
$5) (x-5)(x-1)=(x-1)(x-2)$
$⇔x^2-x-5x+5 =x^2-2x-x+2$
$⇔x^2-x^2-x-5x+2x+x=2-5$
$⇔-3x = -3$
$⇔ x = -3 : (-3)$
$⇔x=1$
Vậy $x=1$
$6) 6(x-3)(x-4) -6x(x-2) =4$
$⇔ 6x^2-25x-18x+72 -6x^2+12x =4$
$⇔ 6x^2-6x^2-25x-18x+12x = 4 -72$
$⇔ -31x = -68$
$⇔ x = \dfrac{68}{31}$
Vậy $x=\dfrac{68}{31}$
$7) -(x+3)(x-4) + (x-1)(x+1) =10$
$⇔ -x^2+4x-3x+12 +x^2-1 =10$
$⇔ -x^2-x^2+4x-3x = 10 +1-12$
$⇔ x = -1$
Vậy $x=-1$
$8) (2x-1)(x-2) - (x+3)(2x-7) = 3$
$⇔ 2x^2-4x-x+2 -2x^2+7x-6x+21 =3$
$⇔ 2x^2-2x^2-4x-x+7x-6x = 3 -21 -2$
$⇔-4x = -20$
$⇔ x = -20 : (-4)$
$⇔ x = 5$
Vậy $x=5$
$9) (x-5)(-x+4) - (x-1)(x+3) =-2x^2$
$⇔ -x^2+4x+5x-20 -x^2+3x+x-3 =-2x^2$
$⇔ -x^2-x^2+2x^2+4x+5x+3x+x = 3 +20$
$⇔13x = 23$
$⇔x = \dfrac{23}{13}$
Vậy $x=-\dfrac{23}{13}$
