Đáp án:
Bài 1: $S=\left\{\dfrac{4}{9}\right\}$
Bài 2: $S=\{0;1\}$
Bài 3: $S=\{2\}$
Giải thích các bước giải:
Bài 1:
ĐKXĐ: $a\ge 0$
$(3\sqrt a-1)(\sqrt a+2)=\sqrt a(2+3\sqrt a)$
$⇔3a+6\sqrt a-\sqrt a-2=2\sqrt a+3a$
$⇔5\sqrt a-2=2\sqrt a$
$⇔3\sqrt a=2$
$⇔9a=4$
$⇔a=\dfrac{4}{9}\,(TM)$
Vậy $S=\left\{\dfrac{4}{9}\right\}$
Bài 2:
ĐKXĐ: $x\in\mathbb R$
$\sqrt{9a^2-6a+1}-a=1$
$⇔\sqrt{(3a-1)^2}=a+1$
$⇔|3a-1|=a+1$
$⇔\left[ \begin{array}{l}3a-1=a=1\\3a-1=-a-1\end{array} \right.⇔\left[ \begin{array}{l}a=1\\a=0\end{array} \right.$
Vậy $S=\{0;1\}$
Bài 3:
ĐKXĐ: $a^3+a^2-3\ge 0$
$\sqrt{a^3+a^2+4}+\sqrt{a^3+a^2-3}=7$
Đặt $t=a^3+a^2\ge 3$
$Pt⇔\sqrt{t+4}+\sqrt{t-3}=7$
$⇔t+4+t-3+2\sqrt{t+4}.\sqrt{t-3}=49$
$⇔2\sqrt{(t+4)(t-3)}=48-2t$
$⇔\sqrt{t^2+t-12}=24-t$
$⇔t^2+t-12=(24-t)^2\,\,(ĐK:\,t\le 24)$
$⇔t^2+t-12=t^2-48t+576$
$⇔49t=588$
$⇔t=12\,(TM)$
$⇔a^3+a^2=12$
$⇔a^3-2a^2+3a^2-6a+6a-12=0$
$⇔(a-2)(a^2+3a+6)=0$
Ta có:
$a^2+3a+6=a^2+2.a.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{15}{4}=\left(a+\dfrac{3}{2}\right)^2+\dfrac{15}{4}>0$
$⇒a-2=0$
$⇒a=2\,(TM)$
Vậy $S=\{2\}$.
