Đáp án:
Câu 1 :
$a/ 2x^2(3x-7x-5)$
$=2x^2(-4x-5)$
$=-8x^3 -10x^2$
$c/ (16x^4 :20x^2y^3 -4x^2y) : (-4x^2)$
$=[16x^4 : (-4x^2)] + [20x^2y^3 : (-4x^2)] + [-4x^2y : (-4x^2)]$
$=-4x^2 -5y^3 +y$
Câu 2 :
$a/ x(x-1) -x^2+2x=5$
$⇔x^2-x-x^2+2x=5$
$⇔x=5$
Vậy $x=5$
$b/ 2x^2-2x=(x-1)^2$
$⇔2x^2-2x =x^2-2x+1$
$⇔2x^2-x^2-2x+2x=1$
$⇔x^2=1$
$⇔x=±1$
$\text{Vậy x ∈ {1 ; -1}}$
$c/ (x+3).(x^2-3x-9) -x(x-2)^2=19$
$⇔x^3 -3x^2-9x +3x^2-9x -27 -x(x^2-4x+4)=19$
$⇔x^3 -18x-27 -x^3 +4x^2-4x =19$
$⇔x^3 -18x-27 -x^3 +4x^2-4x-19=0$
$⇔4x^2-22x -46=0$
$⇔4x^2 -2 . 2x.\dfrac{11}{2} + \dfrac{121}{4} - \dfrac{305}{4} =0$
$⇔(2x-\dfrac{11}{2})^2 -\dfrac{305}{4} =0$
$⇔(2x-\dfrac{11}{2}- \sqrt[]{\dfrac{305}{4}}) .(2x-\dfrac{11}{2} +\sqrt[]{\dfrac{305}{4}})=0$
$⇔$\(\left[ \begin{array}{l}2x-\dfrac{11}{2}-\sqrt[]{\dfrac{305}{4}}=0\\2x-\dfrac{11}{2}+\sqrt[]{\dfrac{305}{4}}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{11+\sqrt[]{305}}{4}\\x=\dfrac{11-\sqrt[]{305}}{4}\end{array} \right.\)
$\text{Vậy x ∈ { $\dfrac{11+\sqrt[]{305}}{4} ; \dfrac{11-\sqrt[]{305}}{4}$ } }$
