Đáp án:
Bài 5 :
$a) (2x+3)(x-4)+(x-5)(x-2) =(3x-5)(x-4)$
$⇔2x² -8x +3x -12 +x^2 -2x -5x +10 = 3x^2 -12x -5x +20$
$⇔ 2x^2 +x^2 -3x^2 -8x +3x -2x -5x +12x +5x = 20 -10 +12$
$⇔ 5x = 22$
$⇔ x = \dfrac{22}{5}$
Vậy....
$b) (8x-3)(3x+2) - (4x+7)(x+4) = (2x+1)(5x-1)$
$⇔ 24x^2 +16x - 9x - 6 -4x^2 -16x -7x -28 = 10x^2 -2x +5x -1$
$⇔ 24x^2 -4x^2 -10x^2 +16x -9x -16x - 7x +2x -5x -6 -28 +1 =0$
$⇔ 10x^2-19x -33 = 0$
$⇔ 10x² -30x + 11x -33 =0$
$⇔ 10x(x-3)+11(x-3)=0$
$⇔ (x-3)(10x+11)=0$
⇔\(\left[ \begin{array}{l}x-3=0\\10x+11=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-\dfrac{11}{10}\end{array} \right.\)
Vậy...
$c) 2x^2 +3(x-1)(x+1) = 5x(x+1)$
$⇔ 2x^2 +3x^2 +3x -3x -3 = 5x^2 +5x$
$⇔ 2x² +3x^2 -5x^2 +3x -3x -5x = 3$
$⇔ -5x=3$
$⇔ x= -\dfrac{3}{5}$
Vậy....
$d) 4(x-1)(x+5) -(x+2)(x+5) =3(x-1)(x+2)$
$⇔ 4x^2+20x -4x -20 - x^2 -5x -2x -10 = 3x^2 +6x -3x +6$
$⇔ 4x^2 -x^2 -3x^2 +20x -4x -5x -2x -6x +3x = 6 +10+20$
$⇔ 6x = 36$
$⇔x = 36 : 6$
$⇔ x = 6$
Vậy ....
Bài 6 :
$a) (a+b+c)(a^2 +b^2+c^2 - ab -bc-ca) = a^3 +b^3 +x^3 -3abc$
VT $ = a^3 +ab^2 +ac^2 -a^2b-abc-ca^2 +a^2b +b^3 +c^2b -ab^2 -b^2c -abc + a^2c +b^2c + c^3 -abc - bc^2-c^2a$
$ = a^3 +b^3 +c^3-3abc $
$\text{⇒đpcm}$
$b)(3a+2b-1)(a+5)-2b(a-2) = (3a+5)(a+3)+2(7b-10)$
$⇔ 3a^2 +15a +2ab +10b -a -5 -2ab +4b = 3a^2 +9a+5a+15 +14b - 20$
$⇔3a^2 +14a +14b-5$ =$ 3a^2 +14a +14b -5$
$\text{⇒ đpcm }$
