Đáp án:
$a,S=\{9\}$
$b, S=\{-2;8\}$
$c,S=\{0;2\}$
$d,S=\left\{1;\dfrac{5}{3}\right\}$
Giải thích các bước giải:
a) ĐKXĐ: $x\ge 5$
$\sqrt{x-5}=2$
$⇔x-5=4$
$⇔x=9\,(TM)$
Vậy $S=\{9\}$
b) ĐKXĐ: $x\in\mathbb R$
$\sqrt{x^2-6x+9}=5$
$⇔\sqrt{(x-3)^2}=5$
$⇔|x-3|=5$
$⇔\left[ \begin{array}{l}x-3=5\\x-3=-5\end{array} \right.⇔\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.$
Vậy $S=\{-2;8\}$
c) ĐKXĐ: $x\in\mathbb R$
$\sqrt{4x^2-4x+1}=x+1$
$⇔\sqrt{(2x-1)^2}=x+1$
$⇔|2x-1|=x+1$
$⇔\left[ \begin{array}{l}2x-1=x+1\\2x-1=-x-1\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=0\end{array} \right.$
Vậy $S=\{0;2\}$
d) ĐKXĐ: $x\in\mathbb R$
$\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}$
$⇔x^2-4x+4=4x^2-12x+9$
$⇔3x^2-8x+5=0$
$⇔(3x-5)(x-1)=0$
$⇔\left[ \begin{array}{l}3x-5=0\\x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=1\end{array} \right.$
Vậy $S=\left\{1;\dfrac{5}{3}\right\}$.
