Toán Lớp 9: A=√x/√x+1 (x≥0)
tìm m để phương trình A=m có nghiệm
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 9: A=√x/√x+1 (x≥0)
tìm m để phương trình A=m có nghiệm
Comments ( 1 )
A = \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
A = m \to \dfrac{{\sqrt x }}{{\sqrt x + 1}} = m\\
\to \sqrt x = m\sqrt x + m\\
\to \left( {m – 1} \right)\sqrt x = – m\\
\to \sqrt x = – \dfrac{m}{{m – 1}}
\end{array}\)
A = \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
A = m \to \dfrac{{\sqrt x }}{{\sqrt x + 1}} = m\\
\to \sqrt x = m\sqrt x + m\\
\to \left( {m – 1} \right)\sqrt x = – m\\
\to \sqrt x = – \dfrac{m}{{m – 1}}\\
\Leftrightarrow \left\{ \begin{array}{l}
– \dfrac{m}{{m – 1}} \ge 0\\
m – 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{m}{{m – 1}} \le 0\\
m \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \ge 0\\
m – 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m \le 0\\
m – 1 > 0
\end{array} \right.
\end{array} \right.\\
m \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \ge 0\\
m < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
m \le 0\\
m > 1
\end{array} \right.\left( l \right)
\end{array} \right.\\
m \ne 1
\end{array} \right.\\
\to 0 \le m < 1
\end{array}\)