√(a+2√(a-1)) + √(a-2√(a-1)) ( với 1≤a≤2) 05/08/2021 Bởi Claire √(a+2√(a-1)) + √(a-2√(a-1)) ( với 1≤a≤2)
Đáp án: $\begin{array}{l}A = \sqrt {a + 2\sqrt {a – 1} } + \sqrt {a – 2\sqrt {a – 1} } \left( {1 \le a \le 2} \right)\\ = \sqrt {a – 1 + 2\sqrt {a – 1} + 1} \\ + \sqrt {a – 1 – 2\sqrt {a – 1} + 1} \\ = \sqrt {{{\left( {\sqrt {a – 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {a – 1} – 1} \right)}^2}} \\ = \sqrt {a – 1} + 1 + \left| {\sqrt {a – 1} – 1} \right|\\ + Do:1 \le a \le 2\\ \Rightarrow 0 \le a – 1 \le 1\\ \Rightarrow 0 \le \sqrt {a – 1} \le 1\\ \Rightarrow \left| {\sqrt {a – 1} – 1} \right| = 1 – \sqrt {a – 1} \\ \Rightarrow A = \sqrt {a – 1} + 1 + 1 – \sqrt {a – 1} = 2\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
A = \sqrt {a + 2\sqrt {a – 1} } + \sqrt {a – 2\sqrt {a – 1} } \left( {1 \le a \le 2} \right)\\
= \sqrt {a – 1 + 2\sqrt {a – 1} + 1} \\
+ \sqrt {a – 1 – 2\sqrt {a – 1} + 1} \\
= \sqrt {{{\left( {\sqrt {a – 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {a – 1} – 1} \right)}^2}} \\
= \sqrt {a – 1} + 1 + \left| {\sqrt {a – 1} – 1} \right|\\
+ Do:1 \le a \le 2\\
\Rightarrow 0 \le a – 1 \le 1\\
\Rightarrow 0 \le \sqrt {a – 1} \le 1\\
\Rightarrow \left| {\sqrt {a – 1} – 1} \right| = 1 – \sqrt {a – 1} \\
\Rightarrow A = \sqrt {a – 1} + 1 + 1 – \sqrt {a – 1} = 2
\end{array}$