Giải các phương trình sau: a) 25$x^{2}$ + 5x - 10x - 2x =0 b) 3x(x+4) - (3x+2)(x - 3) = 100 c) 8x(x - 6) + 7(6 - x) = 0

Đáp án:

$a)$ $\text{S = }$ $\left\{0\:;\:\dfrac{7}{25}\right\}$

$b)$ $\text{S = }$ $\left\{\dfrac{94}{19}\right\}$

$c)$ $\text{S = }$ $\left\{\dfrac{7}{8}\:;\:6\right\}$ 

Giải thích các bước giải:

$a)$ $25x^2+5x-10x-2x=0$

$⇔25x^2-7x=0$

$⇔x(25x−7)=0$

\(⇔\left[ \begin{array}{l}x=0\\25x-7=0\end{array} \right.\) 

\(⇔\left[ \begin{array}{l}x=0\\x=\dfrac{7}{25}\end{array} \right.\) 

$\text{Vậy S = }$ $\left\{0\:;\:\dfrac{7}{25}\right\}$

$b)$ $3x\left(x+4\right)-\left(3x+2\right)\left(x-3\right)=100$

$⇔3x^2+12x-3x^2+9x-2x+6=100$

$⇔12x+9x-2x+6=100$

$⇔19x+6=100$

$⇔19x=100-6$

$⇔19x=94$

$x=\dfrac{94}{19}$

$\text{Vậy S = }$ $\left\{\dfrac{94}{19}\right\}$

$c)$ $8x\left(x\:-\:6\right)\:+\:7\left(6\:-\:x\right)\:=\:0$

$⇔8x^2-48x+42-7x=0$

$⇔8x^2-7x-48x+42=0$

$⇔x\left(8x-7\right)-6\left(8x-7\right)=0$

$⇔\left(8x-7\right)\left(x-6\right)=0$

\(⇔\left[ \begin{array}{l}8x-7=0\quad \\x-6=0\end{array} \right.\) 

\(⇔\left[ \begin{array}{l}x=\dfrac{7}{8}\\x=6\end{array} \right.\)

$\text{Vậy S = }$ $\left\{\dfrac{7}{8}\:;\:6\right\}$