Toán 1,Giải PT sau a,(3x+1)(x-3) ²=(3x+1)(2x-5) ² 04/12/2021 By Skylar 1,Giải PT sau a,(3x+1)(x-3) ²=(3x+1)(2x-5) ²
Đáp án: Giải thích các bước giải: (3x+1)(x-3) ²=(3x+1)(2x-5) ² <=> (3x+1)(x^2-6x+9-4x^2+20x-25)=0 <=> (3x+1)(-3x^2+14x-16)=0 <=> 3x+1=0 <=> 3x=-1 <=> x=-1/3 Vậy x=-1/3 Trả lời
a, $(3x+1)(x-3)^2=(3x+1)(2x-5)^2$ $⇔(3x+1)[(x-3)^2-(2x-5)^2]=0$ $⇔(3x+1)(x-3-2x+5)(x-3+2x-5)=0$ $⇔(3x+1)(-x+2)(3x-8)=0$ $⇔$\(\left[ \begin{array}{l}3x+1=0\\-x+2=0\\3x-8=0\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x=-1/3\\x=2\\x=8/3\end{array} \right.\) Vậy $S=${$-1/3;2;8/3$} Trả lời
Đáp án:
Giải thích các bước giải:
(3x+1)(x-3) ²=(3x+1)(2x-5) ²
<=> (3x+1)(x^2-6x+9-4x^2+20x-25)=0
<=> (3x+1)(-3x^2+14x-16)=0
<=> 3x+1=0
<=> 3x=-1
<=> x=-1/3
Vậy x=-1/3
a, $(3x+1)(x-3)^2=(3x+1)(2x-5)^2$
$⇔(3x+1)[(x-3)^2-(2x-5)^2]=0$
$⇔(3x+1)(x-3-2x+5)(x-3+2x-5)=0$
$⇔(3x+1)(-x+2)(3x-8)=0$
$⇔$\(\left[ \begin{array}{l}3x+1=0\\-x+2=0\\3x-8=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-1/3\\x=2\\x=8/3\end{array} \right.\)
Vậy $S=${$-1/3;2;8/3$}