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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
x^2-2x=0
x.x-2.x=o
x.(x-2)=0
Suy ra x=0 hoặc x-2=0