(x+√(x^2-2x))/(x-√(x^2-2x)) – (x-√(x^2-2x))/(x+√(x^2-2x)) a/tìm ĐKXD b/Rút gọn A c/Tìm x để A<2 06/07/2021 Bởi Clara (x+√(x^2-2x))/(x-√(x^2-2x)) – (x-√(x^2-2x))/(x+√(x^2-2x)) a/tìm ĐKXD b/Rút gọn A c/Tìm x để A<2
Đáp án: $\begin{array}{l}a)A = \dfrac{{x + \sqrt {{x^2} – 2x} }}{{x – \sqrt {{x^2} – 2x} }} – \dfrac{{x – \sqrt {{x^2} – 2x} }}{{x + \sqrt {{x^2} – 2x} }}\\Dkxd:\left\{ \begin{array}{l}{x^2} – 2x \ge 0\\x – \sqrt {{x^2} – 2x} \# 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x\left( {x – 2} \right) \ge 0\\x\# \sqrt {{x^2} – 2x} \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge 2\\x \le 0\end{array} \right.\\{x^2}\# {x^2} – 2x\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge 2\\x \le 0\end{array} \right.\\x\# 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x \ge 2\\x < 0\end{array} \right.\\Vậy\,x < 0\,hoặc\,x \ge 2\\b)A = \dfrac{{x + \sqrt {{x^2} – 2x} }}{{x – \sqrt {{x^2} – 2x} }} – \dfrac{{x – \sqrt {{x^2} – 2x} }}{{x + \sqrt {{x^2} – 2x} }}\\ = \dfrac{{{{\left( {x + \sqrt {{x^2} – 2x} } \right)}^2} – {{\left( {x – \sqrt {{x^2} – 2x} } \right)}^2}}}{{\left( {x + \sqrt {{x^2} – 2x} } \right).\left( {x – \sqrt {{x^2} – 2x} } \right)}}\\ = \dfrac{{{x^2} + 2.x.\sqrt {{x^2} – 2x} + {x^2} – 2x – \left( {{x^2} – 2x\sqrt {{x^2} – 2x} + {x^2} – 2x} \right)}}{{{x^2} – \left( {{x^2} – 2x} \right)}}\\ = \dfrac{{4x\sqrt {{x^2} – 2x} }}{{2x}}\\ = 2\sqrt {{x^2} – 2x} \\c)A < 2\\ \Leftrightarrow 2\sqrt {{x^2} – 2x} < 2\\ \Leftrightarrow \sqrt {{x^2} – 2x} < 1\\ \Leftrightarrow {x^2} – 2x < 1\\ \Leftrightarrow {x^2} – 2x + 1 < 2\\ \Leftrightarrow {\left( {x – 1} \right)^2} < 2\\ \Leftrightarrow – \sqrt 2 < x – 1 < \sqrt 2 \\ \Leftrightarrow 1 – \sqrt 2 < x < 1 + \sqrt 2 \\Do:\left[ \begin{array}{l}x \ge 2\\x < 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2 \le x < 1 + \sqrt 2 \\1 – \sqrt 2 < x < 0\end{array} \right.\\Vậy\,1 – \sqrt 2 < x < 0\,hoặc\,2 \le x < 1 + \sqrt 2 \end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)A = \dfrac{{x + \sqrt {{x^2} – 2x} }}{{x – \sqrt {{x^2} – 2x} }} – \dfrac{{x – \sqrt {{x^2} – 2x} }}{{x + \sqrt {{x^2} – 2x} }}\\
Dkxd:\left\{ \begin{array}{l}
{x^2} – 2x \ge 0\\
x – \sqrt {{x^2} – 2x} \# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {x – 2} \right) \ge 0\\
x\# \sqrt {{x^2} – 2x}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
{x^2}\# {x^2} – 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
x\# 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\\
Vậy\,x < 0\,hoặc\,x \ge 2\\
b)A = \dfrac{{x + \sqrt {{x^2} – 2x} }}{{x – \sqrt {{x^2} – 2x} }} – \dfrac{{x – \sqrt {{x^2} – 2x} }}{{x + \sqrt {{x^2} – 2x} }}\\
= \dfrac{{{{\left( {x + \sqrt {{x^2} – 2x} } \right)}^2} – {{\left( {x – \sqrt {{x^2} – 2x} } \right)}^2}}}{{\left( {x + \sqrt {{x^2} – 2x} } \right).\left( {x – \sqrt {{x^2} – 2x} } \right)}}\\
= \dfrac{{{x^2} + 2.x.\sqrt {{x^2} – 2x} + {x^2} – 2x – \left( {{x^2} – 2x\sqrt {{x^2} – 2x} + {x^2} – 2x} \right)}}{{{x^2} – \left( {{x^2} – 2x} \right)}}\\
= \dfrac{{4x\sqrt {{x^2} – 2x} }}{{2x}}\\
= 2\sqrt {{x^2} – 2x} \\
c)A < 2\\
\Leftrightarrow 2\sqrt {{x^2} – 2x} < 2\\
\Leftrightarrow \sqrt {{x^2} – 2x} < 1\\
\Leftrightarrow {x^2} – 2x < 1\\
\Leftrightarrow {x^2} – 2x + 1 < 2\\
\Leftrightarrow {\left( {x – 1} \right)^2} < 2\\
\Leftrightarrow – \sqrt 2 < x – 1 < \sqrt 2 \\
\Leftrightarrow 1 – \sqrt 2 < x < 1 + \sqrt 2 \\
Do:\left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2 \le x < 1 + \sqrt 2 \\
1 – \sqrt 2 < x < 0
\end{array} \right.\\
Vậy\,1 – \sqrt 2 < x < 0\,hoặc\,2 \le x < 1 + \sqrt 2
\end{array}$