1 giải bt A x mũ 3 – 25x = 0 B x-5/3 + 2 = 4x+1/5 2 nhân các đa thức [x-2].[x+2]= [x-y].[x-y]= [x-1].[x mũ 2 +2x+1]= [x-2].[x mũ 2 – 4x +4]=

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 :

$(x-2)(x+2)= x^2 +2x -2x -4 = x^2-4$

$(x-y)(x-y) = x^2 -xy-xy+y^2 = x^2-2xy+y^2$

$(x-1)(x^2+2x+1) = x^3 +2x^2 +x -x^2-2x -1 = x^3+x^2-x-1$

$(x-2)(x^2-4x+4) = x^3 -4x^2+4x -2x^2+8x-8 = x^3-6x^2+12x-8$