Đáp án:
Nếu $\dfrac{1}{2} \leq x < 1$ thì $P=\sqrt{4x-2}$
Nếu $x \geq 1$ thì $P=\sqrt{2}$
Giải thích các bước giải:
$P=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}$ $(*)$
$\text{ĐKXĐ: $\begin{cases}2x-1 \geq 0 \\x+ \sqrt{2x-1} \geq 0\\x- \sqrt{2x-1} \geq 0\end{cases}$}$
$⇔ \begin{cases}x \geq \dfrac{1}{2} \\x+ \sqrt{2x-1} \geq 0\\x \geq \sqrt{2x-1}\end{cases}$
$⇔ \begin{cases}x \geq \dfrac{1}{2} \\x^2 \geq 2x-1 \text{(do $2x-1 \geq 0$)}\end{cases}$
$⇔ \begin{cases}x \geq \dfrac{1}{2} \\(x-1)^2 \geq 0 \text{(luôn đúng)}\end{cases}$
$⇔ x \geq \dfrac{1}{2}$
$\text{Mặt khác:}$
$(*) ⇔ P^2=(\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}})^2$
$⇔ P^2=x+\sqrt{2x-1}+x-\sqrt{2x-1}-2\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}$
$⇔ P^2=2x-2\sqrt{x^2-(2x-1)}$
$⇔ P^2=2x-2\sqrt{x^2-2x+1}$
$⇔ P^2=2x-2\sqrt{(x-1)^2}$
$⇔ P^2=2x-2|x-1|$
$\text{Nếu $\dfrac{1}{2} \leq x < 1$:}$
$⇔ P^2=2x+2(x-1)$
$⇔ P^2=4x-2$
$⇔ P=\sqrt{4x-2}$
$\text{Nếu $x \geq 1$:}$
$⇔ P^2=2x-2(x-1)$
$⇔ P^2=2$
$⇔ P=\sqrt{2}$
$\text{Vậy nếu $\dfrac{1}{2} \leq x < 1$ thì $P=\sqrt{4x-2}$}$
$\text{Nếu $x \geq 1$ thì $P=\sqrt{2}$}$
