Đáp án:
$\text{ Bài 4 : }$
$a) 4x^2 - 49 = 0$
$⇔ (2x -7)(2x+7)=0$
⇔\(\left[ \begin{array}{l}2x-7=0\\2x+7=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{array} \right.\)
$\text{Vậy x ∈ { $\dfrac{7}{2}$ ; -$\dfrac{7}{2}$ } }$
$b)x² + 36 = 12x$
$⇔x² -12x + 36 =0$
$⇔(x - 6)² = 0$
$⇔ x = 6$
$\text{Vậy x = 6}$
$c) \dfrac{1}{16}x² - x + 4 =0 $
$⇔ (\dfrac{1}{4}X -2)² = 0$
$⇔\dfrac{1}{4}x - 2 =0$
$⇔ \dfrac{1}{4}$ = 2$
$⇔ x =2 : \dfrac{1}{4}$
$⇔ x = 8$
$\text{Vậy x = 8 }$
$d) x³ - 3√3x² + 9x -3√3 = 0$
$⇔ (x - √3)² = 0$
$⇔ x - √3 =0$
$⇔ x = √3$
$\text{Vậy x= √3}$
$e) (x-2)² - 16 =0 $
$⇔ (x-2 -4) (x-2+4) = 0$
$⇔ (x-6)(x+2) = 0$
⇔\(\left[ \begin{array}{l}x-6=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=6\\x=-2\end{array} \right.\)
$\text{Vậy x ∈ { 6 ; - 2} }$
$f) x² -5x -14 = 0$
$⇔ x² +2x -7x -14 = 0$
$⇔ x(x+2)-7(x+2) =0$
$⇔(x+2)(x-7) = 0$
⇔\(\left[ \begin{array}{l}x+2=0\\x-7=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=7\end{array} \right.\)
$\text{Vậy x ∈ { -2 ; 7 } }$
