Đáp án+Giải thích các bước giải:
b,
`\sqrt{4x-8}-12\sqrt{\frac{x-2}{9}}=-1(x≥2)`
`⇔2\sqrt{x-2}-12.\frac{1}{3}\sqrt{x-2}=-1`
`⇔2\sqrt{x-2}-4\sqrt{x-2}=-1`
`⇔-2\sqrt{x-2}=-1`
`⇔\sqrt{x-2}=\frac{1}{2}`
`⇔x-2=\frac{1}{4}`
`⇔x=\frac{9}{4}(tm)`
Vậy `S={\frac{9}{4}}`
c,
`(2\sqrt{x}-1)(\sqrt{x}-2)=7(x≥0)`
`⇔2x-5\sqrt{x}+2=7`
`⇔2x-5\sqrt{x}-5=0`
`⇔2x-5\sqrt{x}+\frac{25}{8}-\frac{65}{8}=0`
`⇔(\sqrt{2x}-\frac{5\sqrt{2}}{4})^2-\frac{65}{8}=0`
`⇔(\sqrt{2x}-\frac{5\sqrt{2}}{4}+\frac{\sqrt{130}}{4})(\sqrt{2x}-\frac{5\sqrt{2}}{4}-\frac{\sqrt{130}}{4})=0`
`⇔(\sqrt{2x}+\frac{\sqrt{130}-5\sqrt{2}}{4})(\sqrt{2x}-\frac{\sqrt{130}+5\sqrt{2}}{4})=0`
$⇔\left[\begin{matrix}\sqrt{2x}+\dfrac{\sqrt{130}-5\sqrt{2}}{4}=0\\\sqrt{2x}-\dfrac{\sqrt{130}+5\sqrt{2}}{4}=0\end{matrix}\right.$
$⇔\left[\begin{matrix}\sqrt{2x}=-\dfrac{\sqrt{130}-5\sqrt{2}}{4}\\\sqrt{2x}=\dfrac{\sqrt{130}+5\sqrt{2}}{4}\end{matrix}\right.$
$⇔\left[\begin{matrix}\sqrt{x}=-\dfrac{\sqrt{65}-5}{4}(L)\\\sqrt{x}=\dfrac{\sqrt{65}+5}{4}\end{matrix}\right.$
$x=\dfrac{45+5\sqrt{65}}{8}(tm)$
Vậy `S={\frac{45+5\sqrt{65}}{8}}`
