Cho x$\geq$1, rút gọn: A= $\sqrt[2]{x+2\sqrt[2]{x-1}}$+$\sqrt[2]{2-2\sqrt[2]{x-1}}$ 11/07/2021 Bởi Reagan Cho x$\geq$1, rút gọn: A= $\sqrt[2]{x+2\sqrt[2]{x-1}}$+$\sqrt[2]{2-2\sqrt[2]{x-1}}$
Đáp án: $A = \left[\begin{array}{l}2\sqrt{x-1}\quad (x \geq 2)\\2\quad \qquad(1 \leq x < 2)\end{array}\right.$ Giải thích các bước giải: $A = \sqrt{x + 2\sqrt{x -1}} + \sqrt{x – 2\sqrt{x -1}}$ $\to A = \sqrt{x-1 + 2\sqrt{x -1}+1} + \sqrt{x-1 – 2\sqrt{x -1}+1}$ $\to A = \sqrt{(\sqrt{x -1} + 1)^2} + \sqrt{(\sqrt{x -1} – 1)^2}$ $\to A = |\sqrt{x -1} + 1| + |\sqrt{x -1} – 1|$ $\to A = \sqrt{x -1} + 1 + |\sqrt{x -1} – 1|$ $\to A = \left[\begin{array}{l}\sqrt{x -1} + 1 + \sqrt{x -1} – 1\quad (x \geq 2)\\\sqrt{x -1} + 1 + 1 – \sqrt{x -1}\quad(1 \leq x < 2)\end{array}\right.$ $\to A = \left[\begin{array}{l}2\sqrt{x-1}\quad (x \geq 2)\\2\quad\qquad (1 \leq x < 2)\end{array}\right.$ Bình luận
Đáp án:
$A = \left[\begin{array}{l}2\sqrt{x-1}\quad (x \geq 2)\\2\quad \qquad(1 \leq x < 2)\end{array}\right.$
Giải thích các bước giải:
$A = \sqrt{x + 2\sqrt{x -1}} + \sqrt{x – 2\sqrt{x -1}}$
$\to A = \sqrt{x-1 + 2\sqrt{x -1}+1} + \sqrt{x-1 – 2\sqrt{x -1}+1}$
$\to A = \sqrt{(\sqrt{x -1} + 1)^2} + \sqrt{(\sqrt{x -1} – 1)^2}$
$\to A = |\sqrt{x -1} + 1| + |\sqrt{x -1} – 1|$
$\to A = \sqrt{x -1} + 1 + |\sqrt{x -1} – 1|$
$\to A = \left[\begin{array}{l}\sqrt{x -1} + 1 + \sqrt{x -1} – 1\quad (x \geq 2)\\\sqrt{x -1} + 1 + 1 – \sqrt{x -1}\quad(1 \leq x < 2)\end{array}\right.$
$\to A = \left[\begin{array}{l}2\sqrt{x-1}\quad (x \geq 2)\\2\quad\qquad (1 \leq x < 2)\end{array}\right.$