Question

a) x^2 - 3x = 0
b)( x + 2)(2x -1) - 3(2x -1) = 0
c) x ( x^2 - 2x +1) + x( -x^2 + 3) = 0

Answer 1

a) x^2 - 3x = 0

    x^2 = 3x

    x^2 : x = 3x : x

    x = 3

b) (x + 2)(2x - 1) - 3(2x - 1) = 0

    (x + 2 - 3)(2x - 1) = 0

    (x - 1)(2x - 1) = 0

Trường hợp1: x - 1= 0

                       x = 1

Trường hợp 2: 2x - 1 = 0

                        2x = 1

                          x = \(\dfrac{1}{2}\)

Vậy x ∈ {1;\(\dfrac{1}{2}\)}

c) x(x^2 - 2x + 1) + x(-x^2 + 3) = 0

   x(x^2 - 2x + 1 -x^2 + 3) = 0

   x(-2x + 4) = 0

Trường hợp 1: x = 0

Trường hợp 2: -2x + 4 = 0

                        -2x = -4

                        x=2

Vậy x  ∈ {0;2}

Answer 2

a. ... \(\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

b. ... \(\Leftrightarrow2x^2+3x-2-6x+3=0\) 

 \(\Leftrightarrow2x^2-3x+1=0\)  \(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

c. ... \(\Leftrightarrow-2x^2+4x=0\Leftrightarrow-x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)