Đáp án+Giải thích các bước giải:
a,
$\sqrt{2x-1}-5=0(x≥\dfrac{1}{2})$
$⇔\sqrt{2x-1}=5$
$⇔2x-1=25$
$⇔2x=26$
$⇔x=13(t/m)$
Vậy `S={13}`
b,
$\dfrac{1}{3}\sqrt{9(2x-1)}-\sqrt{8x-4}+\sqrt{25(2x-1)}=12(x≥\dfrac{1}{2})$
$⇔\sqrt{2x-1}-2\sqrt{2x-1}+5\sqrt{2x-1}=12$
$⇔4\sqrt{2x-1}=12$
$⇔\sqrt{2x-1}=3$
$⇔2x-1=9$
$⇔2x=10$
$⇔x=5(t/m)$
Vậy `S={5}`
c,
$\sqrt{4x^2-4x+1}=x+2$
$⇔|2x-1|=x+2$
$⇔\left[\begin{matrix}2x-1=x+2\\2x-1=-x-2\end{matrix}\right.$
$⇔\left[\begin{matrix}x=3\\3x=-1\end{matrix}\right.$
$⇔\left[\begin{matrix}x=3\\x=\dfrac{-1}{3}\end{matrix}\right.$
Vậy `S={3;\frac{-1}{3}}`
