Toán Rút gọn biểu thức sau: √x- 2√x-2 -1 – √x+2+4√x-2 +3=0 04/08/2021 By Audrey Rút gọn biểu thức sau: √x- 2√x-2 -1 – √x+2+4√x-2 +3=0
Giải thích các bước giải: Ta có: \(\begin{array}{l}A = \sqrt {x – 2\sqrt {x – 2} – 1} – \sqrt {x + 2 + 4\sqrt {x – 2} } + 3\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\ = \sqrt {\left( {x – 2} \right) – 2\sqrt {x – 2} + 1} + \sqrt {\left( {x – 2} \right) + 4\sqrt {x – 2} + 4} + 3\\ = \sqrt {{{\left( {\sqrt {x – 2} – 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x – 2} + 2} \right)}^2}} + 3\\ = \left| {\sqrt {x – 2} – 1} \right| + \left| {\sqrt {x – 2} + 2} \right| + 3\\ = \left| {\sqrt {x – 2} – 1} \right| + \sqrt {x – 2} + 2 + 3\\ = \left| {\sqrt {x – 2} – 1} \right| + \sqrt {x – 2} + 5\\TH1:\,\,\,\,2 \le x \le 3 \Leftrightarrow 0 \le \sqrt {x – 2} \le 1 \Rightarrow \sqrt {x – 2} – 1 \le 0\\ \Rightarrow \left| {\sqrt {x – 2} – 1} \right| = – \left( {\sqrt {x – 2} – 1} \right)\\ \Rightarrow A = – \left( {\sqrt {x – 2} – 1} \right) + \sqrt {x – 2} + 5 = 6\\TH2:\,\,\,x > 3 \Rightarrow \sqrt {x – 2} > 1 \Rightarrow \sqrt {x – 2} – 1 > 0\\ \Rightarrow \left| {\sqrt {x – 2} – 1} \right| = \sqrt {x – 2} – 1\\ \Rightarrow A = \sqrt {x – 2} – 1 + \sqrt {x – 2} + 5 = 2\sqrt {x – 2} + 4\end{array}\) Trả lời
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \sqrt {x – 2\sqrt {x – 2} – 1} – \sqrt {x + 2 + 4\sqrt {x – 2} } + 3\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
= \sqrt {\left( {x – 2} \right) – 2\sqrt {x – 2} + 1} + \sqrt {\left( {x – 2} \right) + 4\sqrt {x – 2} + 4} + 3\\
= \sqrt {{{\left( {\sqrt {x – 2} – 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x – 2} + 2} \right)}^2}} + 3\\
= \left| {\sqrt {x – 2} – 1} \right| + \left| {\sqrt {x – 2} + 2} \right| + 3\\
= \left| {\sqrt {x – 2} – 1} \right| + \sqrt {x – 2} + 2 + 3\\
= \left| {\sqrt {x – 2} – 1} \right| + \sqrt {x – 2} + 5\\
TH1:\,\,\,\,2 \le x \le 3 \Leftrightarrow 0 \le \sqrt {x – 2} \le 1 \Rightarrow \sqrt {x – 2} – 1 \le 0\\
\Rightarrow \left| {\sqrt {x – 2} – 1} \right| = – \left( {\sqrt {x – 2} – 1} \right)\\
\Rightarrow A = – \left( {\sqrt {x – 2} – 1} \right) + \sqrt {x – 2} + 5 = 6\\
TH2:\,\,\,x > 3 \Rightarrow \sqrt {x – 2} > 1 \Rightarrow \sqrt {x – 2} – 1 > 0\\
\Rightarrow \left| {\sqrt {x – 2} – 1} \right| = \sqrt {x – 2} – 1\\
\Rightarrow A = \sqrt {x – 2} – 1 + \sqrt {x – 2} + 5 = 2\sqrt {x – 2} + 4
\end{array}\)