(12x^2-3)(x+3)+(2x^2+7x+3)(x-3)=0 giải pt ạ mng giúp em với 01/12/2021 Bởi Sarah (12x^2-3)(x+3)+(2x^2+7x+3)(x-3)=0 giải pt ạ mng giúp em với
$(12x²-3)(x+3)+(2x²+7x+3)(x-3)=0$ $⇔12x³+36x²-3x-9+2x³-6x²+7x²-21x+3x-9=0$ $⇔14x³+37x²-21x-18=0$ $⇔(14x³+7x²)+(30x²+15x)-(36x+18)=0$ $⇔(2x+1)(7x²+15x-18)=0$ $⇔(2x+1)[(7x²+21x)-(6x+18)]=0$ $⇔(2x+1)(x+3)(7x-6)=0$ $⇔\left[ \begin{array}{l}2x+1=0\\x+3=0\\7x-6=0\end{array} \right.⇔\left[ \begin{array}{l}x=\frac{-1}{2}\\x=-3\\x=\frac{6}{7}\end{array} \right.$ Vậy $S=${$\frac{-1}{2};-3;\frac{6}{7}$}. Bình luận
Đáp án: Giải thích các bước giải: $(12x² – 3)(x + 3) + (2x² + 7x + 3)(x – 3) = 0$ $⇔(12x² – 6x + 6x – 3)(x + 3)+(2x² + 6x + x + 3)(x – 3) = 0$ $⇔ [2x(6x – 3) + (6x – 3)](x + 3)+[2x(x + 3) + (x + 3)](x – 3) = 0$ $⇔(2x + 1)(x + 3)(6x – 3) + (2x + 1)(x + 3)(x – 3) = 0$ $⇔ (2x + 1)(x + 3)[(6x – 3) + (x – 3)] = 0$ $⇔ (2x + 1)(x + 3)(7x – 6) = 0$ $ ⇒ x = -\frac{1}{2}; x = – 3; x = \frac{6}{7}$ Bình luận
$(12x²-3)(x+3)+(2x²+7x+3)(x-3)=0$
$⇔12x³+36x²-3x-9+2x³-6x²+7x²-21x+3x-9=0$
$⇔14x³+37x²-21x-18=0$
$⇔(14x³+7x²)+(30x²+15x)-(36x+18)=0$
$⇔(2x+1)(7x²+15x-18)=0$
$⇔(2x+1)[(7x²+21x)-(6x+18)]=0$
$⇔(2x+1)(x+3)(7x-6)=0$
$⇔\left[ \begin{array}{l}2x+1=0\\x+3=0\\7x-6=0\end{array} \right.⇔\left[ \begin{array}{l}x=\frac{-1}{2}\\x=-3\\x=\frac{6}{7}\end{array} \right.$
Vậy $S=${$\frac{-1}{2};-3;\frac{6}{7}$}.
Đáp án:
Giải thích các bước giải:
$(12x² – 3)(x + 3) + (2x² + 7x + 3)(x – 3) = 0$
$⇔(12x² – 6x + 6x – 3)(x + 3)+(2x² + 6x + x + 3)(x – 3) = 0$
$⇔ [2x(6x – 3) + (6x – 3)](x + 3)+[2x(x + 3) + (x + 3)](x – 3) = 0$
$⇔(2x + 1)(x + 3)(6x – 3) + (2x + 1)(x + 3)(x – 3) = 0$
$⇔ (2x + 1)(x + 3)[(6x – 3) + (x – 3)] = 0$
$⇔ (2x + 1)(x + 3)(7x – 6) = 0$
$ ⇒ x = -\frac{1}{2}; x = – 3; x = \frac{6}{7}$