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Trả lời:
$1,K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}$
$⇒K^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}$
$=6+2\sqrt{9-5}$
$=6+2.2=10$
$⇒K=\sqrt{10}$
$2, \sqrt{x-3}-2.\sqrt{x^2-3x}=0$ $(ĐK:\,x\ge 3)$
$⇔\sqrt{x-3}.(1-2\sqrt{x})=0$
$⇔\left[ \begin{array}{l}\sqrt{x-3}=0\\2\sqrt{x}=1\end{array} \right.⇔\left[ \begin{array}{l}x=3\\x=\dfrac{1}{4}\,(L)\end{array} \right.$
Vậy `S={3}`.
$3,\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}$ $\bigg{(}ĐK:\,x>-\dfrac{5}{7}\bigg{)}$
$⇔9x-7=7x+5$
$⇔2x=12$
$⇔x=6\,(TM)$
Vậy $S=\{6\}$.
$4,x-5\sqrt{x}+4=0$ $(ĐK:x\ge 0)$
$⇔(\sqrt{x}-1)(\sqrt{x}-4)=0$
$⇔\left[ \begin{array}{l}x=1\\x=16\end{array} \right.$
Vậy $S=\{1;16\}$.
