Bài 3a)

\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)

mà \(a+b=-c\Rightarrow a^3+b^3+c^3=3abc\)

\(3x\left(x^2-4\right)=0\)

\(3x\left(x-2\right)\left(x+2\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+2=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)

\(2x^2-x-6=0\)

\(2x^2-4x+3x-6=0\)

\(2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\left(x-2\right)\left(2x+3\right)=0\)

\(\left[\begin{array}{nghiempt}x-2=0\\2x+3=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=-\frac{3}{2}\end{array}\right.\)

1) \(3x(x^2-4)=0 \)

\(=> 3x(x-2)(x+2)=0\)

\(=>\left[\begin{array}{nghiempt}3x=0\\x-2=0\\x+2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)

Vậy \(x\in\left\{0;2;-2\right\}\)

2) \(2x^2-x-6=0\)

\(2x^2-4x+3x-6=0\)

\(\left(2x^2-4x\right)+\left(3x-6\right)=0\)

\(2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\left(x-2\right)\left(2x+3\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\2x+3=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{3}{2}\end{array}\right.\)

Vậy \(x\in\left\{2;-\frac{3}{2}\right\}\)

1. <=> (x-2).(2x+3) = 0

<=> x-2=0 hoặc 2x+3 = 0

<=> x=2 hoặc x=-3/2

2. <=> x^2-4x+4-x^2+9 = 0

<=> 13-4x=0

<=> 4x=13

<=> x = 13/4

3.<=>4x^2-24x+36 - 4x^2+1 =  10

<=> 37-24x = 10

<=> 24x = 37 - 10 = 27

<=> x = 27 : 24 = 9/8

k mk nha

2. \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(x^2-4x+4-x^2+9=0\)

\(-4x+13=0\)

\(-4x=-13\)

\(x=\frac{13}{4}\)

vậy \(x=\frac{13}{4}\)