Đáp án+Giải thích các bước giải:
a,
$\sqrt{2x-1}=4(x≥\dfrac{1}{2})$
$⇔2x-1=16$
$⇔2x=17$
$⇔x=\dfrac{17}{2}(t/m)$
Vậy `S={\frac{17}{2}}`
b,
`\sqrt{4x^2-4x+1}=\sqrt{3}-1`
`⇔4x^2-4x+1=(\sqrt{3}-1)^2`
`⇔(2x-1)^2-(\sqrt{3}-1)^2=0`
`⇔(2x-1+\sqrt{3}-1)(2x-1-\sqrt{3}+1)=0`
`⇔(2x+\sqrt{3}-2)(2x-\sqrt{3})=0`
$⇔\left[\begin{matrix}2x+\sqrt{3}-2=0\\2x-\sqrt{3}=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\dfrac{2-\sqrt{3}}{2}\\x=\dfrac{\sqrt{3}}{2}\end{matrix}\right.$
Vậy `S={\frac{2-\sqrt{3}}{2};\frac{\sqrt{3}}{2}}`
