Câu hỏi của Kiều Chinh

Question

Tìm MinA: $\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}$

TFBoys · Accepted Answer

ĐK: $x\ge1$

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

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

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

$\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2$

Đẳng thức xảy ra $\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0$

$\Leftrightarrow1\le x\le2$Câu trả lời: ĐK: $x\ge1$

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

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

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

$\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2$

Đẳng thức xảy ra $\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0$

$\Leftrightarrow1\le x\le2$

Bài 8: Rút gọn biểu thức chứa căn bậc hai