tìm x biết: a)2x(x-5)-x(3+2x)=26 b)5x(1-2x)-3x(x+18)=0 c)5x-3{4x-2[4x-3(5x-2)]}=182 d)(x+2)(x+3)-(x-2)(x+5)=0 e)(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-

Đáp án:

 $a) 2x(x-5) -x(3+2x)=26$

$⇔ 2x^2-10x -3x-2x^2 =26$

$⇔ -13x=26$

$⇔x = 26 : (-13)$

$⇔x = -2 $

Vậy $x=-2$

$b) 5x(1-2x) -3x(x+18) =0$

$⇔ 5x-10x^2-3x^2-54x = 0$

$⇔ -13x^2-49x=0$

$⇔x(-13x-49)=0$

⇔\(\left[ \begin{array}{l}x=0\\-13x-49=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=-\dfrac{49}{13}\end{array} \right.\) 

Vậy $\text{ x ∈ { $ 0 ; -\dfrac{49}{13}$ }}$

$c) 5x-3{4x-2) [4x-3(5x-2)]} = 182$

$⇔ 5x-3{4x-2[4x-15x+6]} = 182$

$⇔ 5x -3{4x-8x+30x-12} =182$

$⇔ 5x -12x+24x-90x+36 = 182$

$⇔ 5x-12x+24x-90x = 182-36$

$⇔-73x = 146$

$⇔ x = 146 : (-73)$

$⇔x= -2$

Vậy $x=-2$

$d) (x+2)(x+3) -(x-2)(x+5)=0$

$⇔x^2+3x+2x+6 -x^2-5x+2x+10 =0$

$⇔ 2x +16=0$

$⇔ 2x = -16$

$⇔x=-16 : 2$

$⇔ x = -8$

Vậy $x=-8$

$e) (8x-3)(3x+2) -(4x+7)(x+4) =(2x+1)(5x-1) -33$

$⇔ 24x^2+16x-9x-6 -4x^2-16x-7x-28 = 10x^2-2x+5x-1 -33$

$⇔ 24x^2-4x^2-10x^2 +16x-9x-16x-7x+2x-5x-6-28+1+33=0$

$⇔10x^2-19x=0$

$⇔x(10x-19)=0$

⇔\(\left[ \begin{array}{l}x=0\\10x-19=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=\dfrac{19}{10}\end{array} \right.\) 

Vậy $\text{ x ∈ { $0 ; \dfrac{19}{10}$}}$

 

Trả lời