Đáp án:
$a) x^2 -81=0$
$⇔(x-9)(x+9)=0$
$⇔$\(\left[ \begin{array}{l}x-9=0\\x+9=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=9\\x=-9\end{array} \right.\)
Vậy $\text{x ∈ {9 ; -9} }$
$b) 2x^2 -162=0$
$⇔2x^2 =162$
$⇔x^2 = 162 : 2$
$⇔x^2 =81$
$⇔x^2 = (±9)^2$
$⇔x=±9$
Vậy $\text{x ∈ {9 ; -9} }$
$c) (x+3)^2 =4$
$⇔(x+3)^2 =2^2$
$⇔x+3=2$
$⇔x=2-3$
$⇔x=-1$
Vậy $x=-1$
$d) (x-2)^2 -100=0$
$⇔(x-2-10)(x-2+10)=0$
$⇔(x-12)(x+8)=0$
$⇔$\(\left[ \begin{array}{l}x-12=0\\x+8=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=12\\x=-8\end{array} \right.\)
Vậy $\text{x ∈ {12 ; -8}}$
$e) x^3 =4x$
$⇔x^3-4x=0$
$⇔x(x^2-4)=0$
$⇔x(x-2)(x+2)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x-2=0\\x+2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=0\\x=2\\x=-2\end{array} \right.\)
Vậy $\text{x ∈ {0; -2 ; 2}}$
$f) x^3 -49x=0$
$⇔x(x^2-49)=0$
$⇔x(x-7)(x+7)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x-7=0\\x+7=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=0\\x=7\\x=-7\end{array} \right.\)
Vậy $\text{x ∈ {0 ; -7 ;7}}$
