Đáp án:
$a)3\sqrt{2x-3}-6=9$ $(x≥\dfrac{3}{2})$
$⇔ 3\sqrt{2x-3}=15$
$⇔\sqrt{2x-3}=5$
$⇔2x-3=25$
$⇔2x=28$
$⇔x=14(t/m)$
$b)\sqrt{2}.x^2-\sqrt{98}=0$
$⇔\sqrt{2}.x^2=\sqrt{98}$
$⇔x^2=\sqrt{98}÷\sqrt{2}$
$⇔x^2=7$
$⇔\left[ \begin{array}{l}x=\sqrt{7}\\x=-\sqrt{7}\end{array} \right.$
$c)\sqrt{x^2-9}+\sqrt{x-3}=0$ $(x≥3)$
$⇔\sqrt{(x-3)(x+3)}+\sqrt{x-3}=0$
$⇔ \sqrt{(x-3)}( \sqrt{x+3}+1)=0$
$⇔\left[ \begin{array}{l}\sqrt{(x-3)}=0\\\sqrt{x+3}+1=0(loai)\end{array} \right.$
$⇒x=3(t/m)$
$d)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27\sqrt{\dfrac{x-1}{81}}=4 $ $(x≥1)$
$⇔\dfrac{2}{3}\sqrt{9(x-1)}-\dfrac{1}{4}\sqrt{16(x-1)}+27\sqrt{\dfrac{x-1}{9^2}}=4 $
$⇔\dfrac{2}{3}.3\sqrt{x-1}-\dfrac{1}{4}.4\sqrt{x-1}+27.\dfrac{1}{9}\sqrt{x-1}=4 $
$⇔2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4$
$⇔4\sqrt{x-1}=4$
$⇔\sqrt{x-1}=1$
$⇔x-1=1$
$⇔x=2(t/m)$
