Rút gọn √(x+2√(2x-4)) + √(x-2√(2x-4)) . Mình Sẽ vote 5* cho bạn nào trả lời chính xác nhất. 06/07/2021 Bởi Valentina Rút gọn √(x+2√(2x-4)) + √(x-2√(2x-4)) . Mình Sẽ vote 5* cho bạn nào trả lời chính xác nhất.
ĐK: $x\ge 2$ $I=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ $=\sqrt{x+2.\sqrt2.\sqrt{x-2}}+\sqrt{x-2\sqrt{2}.\sqrt{x-2}}$ $=\sqrt{x-2+2.\sqrt2.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt2.\sqrt{x-2}+2}$ $=\sqrt{ (\sqrt{x-2})^2+2.\sqrt{x-2}.\sqrt2+(\sqrt2)^2}+\sqrt{ (\sqrt{x-2})^2-2.\sqrt{x-2}.\sqrt2+(\sqrt2)^2}$ $=\sqrt{ (\sqrt{x-2}+\sqrt2)^2}+\sqrt{(\sqrt{x-2}-\sqrt2)^2}$ $=|\sqrt{x-2}+\sqrt2|+|\sqrt{x-2}-\sqrt2|$ • Với $\sqrt{x-2}\ge \sqrt2 \to x\ge 4$: $I=\sqrt{x-2}+\sqrt2+\sqrt{x-2}-\sqrt2$ $=2\sqrt{x-2}$ • Với $\sqrt{x-2}\le \sqrt2\to x\le 4\to 2\le x\le 4$: $I=\sqrt{x-2}+\sqrt2-\sqrt{x-2}+\sqrt2$ $=2\sqrt2$ Bình luận
Đáp án: `\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}(x>=2)` `=\sqrt{(2x+4\sqrt{2x-4})/2}+\sqrt{(2x-4\sqrt{2x-4})/2}` `=\sqrt{(2x-4+4\sqrt{2x-4}+4)/2}+\sqrt{(2x-4-4\sqrt{2x-4}+4)/2}` `=\sqrt{(\sqrt{2x-4}+2)^2/2}+\sqrt{(\sqrt{2x-4}+2)^2/2}` `=|\sqrt{2x-4}+2|/\sqrt2+|\sqrt{2x-4}-2|/\sqrt2` `=(\sqrt{2x-4}+2+|\sqrt{2x-4}-2|)/\sqrt2` Bình luận
ĐK: $x\ge 2$
$I=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$
$=\sqrt{x+2.\sqrt2.\sqrt{x-2}}+\sqrt{x-2\sqrt{2}.\sqrt{x-2}}$
$=\sqrt{x-2+2.\sqrt2.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt2.\sqrt{x-2}+2}$
$=\sqrt{ (\sqrt{x-2})^2+2.\sqrt{x-2}.\sqrt2+(\sqrt2)^2}+\sqrt{ (\sqrt{x-2})^2-2.\sqrt{x-2}.\sqrt2+(\sqrt2)^2}$
$=\sqrt{ (\sqrt{x-2}+\sqrt2)^2}+\sqrt{(\sqrt{x-2}-\sqrt2)^2}$
$=|\sqrt{x-2}+\sqrt2|+|\sqrt{x-2}-\sqrt2|$
• Với $\sqrt{x-2}\ge \sqrt2 \to x\ge 4$:
$I=\sqrt{x-2}+\sqrt2+\sqrt{x-2}-\sqrt2$
$=2\sqrt{x-2}$
• Với $\sqrt{x-2}\le \sqrt2\to x\le 4\to 2\le x\le 4$:
$I=\sqrt{x-2}+\sqrt2-\sqrt{x-2}+\sqrt2$
$=2\sqrt2$
Đáp án:
`\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}(x>=2)`
`=\sqrt{(2x+4\sqrt{2x-4})/2}+\sqrt{(2x-4\sqrt{2x-4})/2}`
`=\sqrt{(2x-4+4\sqrt{2x-4}+4)/2}+\sqrt{(2x-4-4\sqrt{2x-4}+4)/2}`
`=\sqrt{(\sqrt{2x-4}+2)^2/2}+\sqrt{(\sqrt{2x-4}+2)^2/2}`
`=|\sqrt{2x-4}+2|/\sqrt2+|\sqrt{2x-4}-2|/\sqrt2`
`=(\sqrt{2x-4}+2+|\sqrt{2x-4}-2|)/\sqrt2`