Đáp án:
$a) 64x^2 -49 =0$
$⇔ (8x-7)(8x+7)=0$
⇔\(\left[ \begin{array}{l}8x-7=0\\8x+7=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{7}{8}\\x=-\dfrac{7}{8}\end{array} \right.\)
Vậy...
$b) (x^2+6x+9) . (x^2+8x+7)=0$
$⇔(x+3)^2 . (x^2+x+7x+7)=0$
$⇔(x+3)^2 . [x(x+1)+7(x+1)]=0$
$⇔(x+3)^2 . (x+1)(x+7)=0$
⇔\(\left[ \begin{array}{l}(x+3)^2=0\\x+1=0\\x+7=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=-1\\x=-7\end{array} \right.\)
Vậy....
$c) (x^2+1).(x^2-8x-20)=0$
Vì $x^2 ≥ 0 ⇔x^2 +1 > 0$ (loại)
$⇔x^2-8x-20=0$
$⇔x^2+2x-10x-20=0$
$⇔x(x+2)-10(x+2)=0$
$⇔(x+2)(x-10)=0$
⇔\(\left[ \begin{array}{l}x+2=0\\x-10=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=10\end{array} \right.\)
Vậy....
