Đáp án:
Câu 1 :
$Ax^3-25x=0$
$⇔ x(x^2-25)=0$
$⇔x(x-5)(x+5)=0$
⇔\(\left[ \begin{array}{l}x=0\\x-5=0\\x+5=0\\\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=5\\x=-5\end{array} \right.\)
Vậy $\text{ x ∈ { $0 ;5 ; -5$ }}$
$B = x-\dfrac{5}{3} +2 = 4x +\dfrac{1}{5}$
$⇔ x -4x = \dfrac{1}{5} -2 +\dfrac{5}{3}$
$⇔ -3x= -\dfrac{2}{15}$
$⇔x = -\dfrac{2}{15} : (-3)$
$⇔x = \dfrac{2}{45}$
Vậy $\text{ x = $\dfrac{2}{15}$}$
Câu 2 :
$(x-2)(x+2)= x^2 +2x -2x -4 = x^2-4$
$(x-y)(x-y) = x^2 -xy-xy+y^2 = x^2-2xy+y^2$
$(x-1)(x^2+2x+1) = x^3 +2x^2 +x -x^2-2x -1 = x^3+x^2-x-1$
$(x-2)(x^2-4x+4) = x^3 -4x^2+4x -2x^2+8x-8 = x^3-6x^2+12x-8$
