a,
7 - 9x + 2x² = 0
⇔ 2x² - 2x - 7x + 7 = 0
⇔ (2x² - 2x ) - ( 7x - 7 ) = 0
⇔ 2x ( x - 1 ) - 7 ( x - 1 ) = 0
⇔ ( x - 1 )(2x - 7 ) = 0
⇔ \(\left[ \begin{array}{l}x=1\\x=7/2\end{array} \right.\)
Vậy x ∈ { 1 ; 7/2}
b,4x^3-4x=0
⇔ 4x(x² - 1 ) = 0
⇔ \(\left[ \begin{array}{l}x=0\\x^2=1\end{array} \right.\)
⇔ x ∈ { 0 ; 1 ; - 1 }
Vậy x ∈ { 0 ; 1 ; - 1 }
