3(x+1)^2-x-1=0 Ai làm giúp với nhanh nha 04/07/2021 Bởi Claire 3(x+1)^2-x-1=0 Ai làm giúp với nhanh nha
Đáp án: \(\left[ \begin{array}{l}x=-\frac{2}{3}\\x=-1\end{array} \right.\) Giải thích các bước giải: $3.(x+1)^2-x-1=0$ $⇔3.(x^2+2x+1)-x-1=0$ $⇔3x^2+6x+3-x-1=0$ $⇔3x^2+5x+2=0$ $⇔3x^2+3x+2x+2=0$ $⇔3x.(x+1)+2.(x+1)=0$ $⇔(3x+2).(x+1)=0$ \(\left[ \begin{array}{l}3x+2=0\\x+1=0\end{array} \right.\) \(\left[ \begin{array}{l}x=-\frac{2}{3}\\x=-1\end{array} \right.\) Bình luận
3(x+1)^2-x-1=0 <=>3(x^2+2x+1)-x-1=0 <=>3x^2+6x+3-x-1=0 <=>3x^2+5x+2=0 <=>3x^2+3x+2x+2=0 <=>(3x^2+3x)+(2x+2)=0 <=>3x(x+1)+2(x+1)=0 <=>(3x+2)(x+1)=0 <=>3x+2=0 x+1=0 <=>3x=-2 x=-1 <=>x=-2/3 x=-1 Bình luận
Đáp án:
\(\left[ \begin{array}{l}x=-\frac{2}{3}\\x=-1\end{array} \right.\)
Giải thích các bước giải:
$3.(x+1)^2-x-1=0$
$⇔3.(x^2+2x+1)-x-1=0$
$⇔3x^2+6x+3-x-1=0$
$⇔3x^2+5x+2=0$
$⇔3x^2+3x+2x+2=0$
$⇔3x.(x+1)+2.(x+1)=0$
$⇔(3x+2).(x+1)=0$
\(\left[ \begin{array}{l}3x+2=0\\x+1=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=-\frac{2}{3}\\x=-1\end{array} \right.\)
3(x+1)^2-x-1=0
<=>3(x^2+2x+1)-x-1=0
<=>3x^2+6x+3-x-1=0
<=>3x^2+5x+2=0
<=>3x^2+3x+2x+2=0
<=>(3x^2+3x)+(2x+2)=0
<=>3x(x+1)+2(x+1)=0
<=>(3x+2)(x+1)=0
<=>3x+2=0
x+1=0
<=>3x=-2
x=-1
<=>x=-2/3
x=-1