Câu hỏi của AEri Sone

Question

Tìm x ;

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

Luân Đào · Accepted Answer

Sửa lại tí:$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}$

ĐK: $x\ge1$

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

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

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

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

Vì $x\ge1\Rightarrow\sqrt{2x-1}-1\ge\sqrt{2-1}-1=0\Rightarrow\left|\sqrt{2x-1}-1\right|=\sqrt{2x-1}-1$

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

$\Leftrightarrow2\sqrt{2x-1}=2$

$\Leftrightarrow\sqrt{2x-1}=1$

$\Leftrightarrow2x-1=1$

$\Leftrightarrow2x=2$

$\Leftrightarrow x=1$ (thỏa)

Vậy S = {1}Câu trả lời: Sửa lại tí:$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}$