Đáp án:
$ 1) (x-5)(x+4) -x(x-11)=5$
$⇒ x²+4x-5x-20 - x² + 11x = 5$
$⇒ x² - x² + 4x - 5x +11x= 5 + 20 $
$⇒ 10x = 25 $
$ ⇒x = 25 : 10$
$⇒ x = 2,5 $
$Vậy x = 2,5$
$2) (x-1)(x+1)-(x-2)(x+4) = 10$
$⇒ x² +x-x-1 - x²-4x+2x+8 = 10$
$ ⇒ x² - x² + x - x - 4x + 2x = 10 +1 - 8$
$⇒ -2x = 3$
$⇒ x = -\dfrac{3}{2}$
$Vậy x = -\dfrac{3}{2}$
$3) (4x²-1)(x+3) - (2x²-1)(2x+1)=x²$
$ ⇒ 4x³ +12x² -x -3 -4x³ -2x²+2x+1 = x²$
$⇒ 4x³ -4x³ +12x² -2x² -x²-x+2x -3 +1 = 0$
$⇒ 9x²+x -2 = 0$
$⇒ (3x)² + 2 . 3x . \dfrac{1}{6} + \dfrac{1}{36} - \dfrac{73}{36} =0 $
$⇒(3x + \dfrac{1}{6})² - \dfrac{73}{36} = 0$
$⇒ (3x + \dfrac{1}{6} + \sqrt[]{\dfrac{73}{36}}) .(3x + \dfrac{1}{6} - \sqrt[]{\dfrac{73}{36}}) = 0$
⇒\(\left[ \begin{array}{l}3x + \dfrac{1}{6} + \sqrt[]{\dfrac{73}{36}}=0\\3x + \dfrac{1}{6} - \sqrt[]{\dfrac{73}{36}}=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=\dfrac{-1-\sqrt[]{73}}{18}\\x=\dfrac{-1+\sqrt[]{73}}{18}\end{array} \right.\)
Vậy ....
$4) (2x-3)(3x+5)+3 =6x(x+2)$
$⇒ 6x² +10x -9x -15 + 3 = 6x² +12x$
$⇒ 6x²-6x² +10x -9x -12x = -3 +15$
$⇒ -11x = 12$
$⇒ x = -\dfrac{12}{11}$
$ Vậy x = -\dfrac{12}{11}$
$5) 2x(x-4) +2x = (x+5)(2x-4)+5$
$⇒ 2x² -8x +2x = 2x² -4x +10x -20 +5$
$⇒ 2x² -2x² -8x +2x +4x -10x = -20 +5 $
$⇒ -12x = -15$
$⇒ x = \dfrac{15}{12}$
$ Vậy x = \dfrac{15}{12}$
$6) (x²+x+1)(x+1) -2x = x²(x+2)+2(x+3)$
$⇒ x³ +x²+x²+x+x+1 -2x = x³ +2x² +2x+6$
$⇒ x³ -x³ +x² +x² -2x² +x +x -2x -2x = 6 -1 $
$⇒ -2x = 5 $
$⇒ x = -\dfrac{5}{2}$
$Vậy x = -\dfrac{5}{2}$
