Question

a, 6x^3-6x=0

b, 2x(3x+7) -6x^2=28

c, 2(4x+4)-5(x-3)=0

Answer 1

a) \(6x^3-6x=0\Leftrightarrow6x\left(x^2-1\right)=0\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)b) \(2x\left(3x+7\right)-6x^2=28\Leftrightarrow6x^2+14x-6x^2=28\Leftrightarrow14x=28\Leftrightarrow x=2\)

c) \(2\left(4x+4\right)-5\left(x-3\right)=0\Leftrightarrow8x+8-5x+15=0\Leftrightarrow3x=-23\Leftrightarrow x=-\dfrac{23}{3}\)

Answer 2

Answer 3

a: Ta có: \(6x^3-6x=0\)

\(\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)

b: Ta có: \(2x\left(3x+7\right)-6x^2=28\)

\(\Leftrightarrow6x^2+14x-6x^2=28\)

\(\Leftrightarrow14x=28\)

hay x=2

c: Ta có: \(2\left(4x+4\right)-5\left(x-3\right)=0\)

\(\Leftrightarrow8x+8-5x+15=0\)

\(\Leftrightarrow3x=-23\)

hay \(x=-\dfrac{23}{3}\)