Đáp án:
Giải thích các bước giải:
$5.x^2-x+\dfrac{1}{4}=0$
$⇔\left (x-\dfrac{1}{2} \right )^2=0$
$⇔x-\dfrac{1}{2}=0$
$⇔x=\dfrac{1}{2}$
$6.x^2+x+\dfrac{1}{4}=0$
$⇔\left (x+\dfrac{1}{2} \right )^2=0$
$⇔x+\dfrac{1}{2}=0$
$⇔x=-\dfrac{1}{2}$
$7.(x+1)^2-(2x+3)^2=0$
$⇔(x+1-2x-3)(x+1+2x+3)=0$
$⇔(-x-2)(3x+4)=0$
⇔\(\left[ \begin{array}{l}-x-2=0\\3x+4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=-\dfrac{4}{3}\end{array} \right.\)
$8.(2x+3)^2=(x-1)^2$
$⇔(2x+3)^2-(x-1)^2=0$
$⇔(2x+3-x+1)(2x+3+x-1)=0$
$⇔(x+4)(3x+2)=0$
⇔\(\left[ \begin{array}{l}x+4=0\\3x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-4\\x=-\dfrac{2}{3}\end{array} \right.\)
Chúc em học tốt.
