Câu hỏi của Big City Boy

Question

Rút gọn biểu thức: $\sqrt{4x-2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}$ khi $x\ge\dfrac{1}{4}$

Nguyễn Việt Lâm · Accepted Answer

$=\sqrt{4x-1-2\sqrt{4x-1}+1}+\sqrt{4x-1+2\sqrt{4x-1}+1}$

$=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}$

$=\left|\sqrt{4x-1}-1\right|+\sqrt{4x-1}+1$

$=\left[{}\begin{matrix}2\sqrt{4x-1}\text{ nếu }x\ge\dfrac{1}{2}\2\text{ nếu }\dfrac{1}{4}\le x< \dfrac{1}{2}\end{matrix}\right.$

Câu trả lời:

$=\sqrt{4x-1-2\sqrt{4x-1}+1}+\sqrt{4x-1+2\sqrt{4x-1}+1}$

$=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}$

$=\left|\sqrt{4x-1}-1\right|+\sqrt{4x-1}+1$

$=\left[{}\begin{matrix}2\sqrt{4x-1}\text{ nếu }x\ge\dfrac{1}{2}\2\text{ nếu }\dfrac{1}{4}\le x< \dfrac{1}{2}\end{matrix}\right.$