Toán Rút gọn $P=\frac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}$ 15/09/2021 By Brielle Rút gọn $P=\frac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}$
Đáp án: $\begin{array}{l}Dkxd:x \ge 1\\P = \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {x – 2\sqrt {x – 1} } }}\\ = \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {x – 1 – 2\sqrt {x – 1} + 1} }}\\ = \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {{{\left( {\sqrt {x – 1} – 1} \right)}^2}} }}\\ = \dfrac{{1 – \sqrt {x – 1} }}{{\left| {\sqrt {x – 1} – 1} \right|}}\\ + Khi:\sqrt {x – 1} – 1 > 0\\ \Rightarrow \sqrt {x – 1} > 1\\ \Rightarrow x – 1 > 1\\ \Rightarrow x > 2\\ \Rightarrow P = \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {x – 1} – 1}} = – 1\\ + Khi:\sqrt {x – 1} – 1 < 0\\ \Rightarrow x < 2\\ \Rightarrow 1 \le x < 2\\ \Rightarrow P = \dfrac{{1 – \sqrt {x – 1} }}{{1 – \sqrt {x – 1} }} = 1\\Vậy\,\left[ \begin{array}{l}P = 1\,khi:x > 2\\P = – 1\,khi:1 \le x < 2\end{array} \right.\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
Dkxd:x \ge 1\\
P = \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {x – 2\sqrt {x – 1} } }}\\
= \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {x – 1 – 2\sqrt {x – 1} + 1} }}\\
= \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {{{\left( {\sqrt {x – 1} – 1} \right)}^2}} }}\\
= \dfrac{{1 – \sqrt {x – 1} }}{{\left| {\sqrt {x – 1} – 1} \right|}}\\
+ Khi:\sqrt {x – 1} – 1 > 0\\
\Rightarrow \sqrt {x – 1} > 1\\
\Rightarrow x – 1 > 1\\
\Rightarrow x > 2\\
\Rightarrow P = \dfrac{{1 – \sqrt {x – 1} }}{{\sqrt {x – 1} – 1}} = – 1\\
+ Khi:\sqrt {x – 1} – 1 < 0\\
\Rightarrow x < 2\\
\Rightarrow 1 \le x < 2\\
\Rightarrow P = \dfrac{{1 – \sqrt {x – 1} }}{{1 – \sqrt {x – 1} }} = 1\\
Vậy\,\left[ \begin{array}{l}
P = 1\,khi:x > 2\\
P = – 1\,khi:1 \le x < 2
\end{array} \right.
\end{array}$