Toán Lớp 9: Giải các bất phương trình sau:
a) $\frac{x – 2}{x – 3}$ > 0
b) $\frac{x + 2 }{x – 5}$ < 0
c) $\frac{x - 1}{x + 3}$ > 1
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: Giải các bất phương trình sau:
a) $\frac{x – 2}{x – 3}$ > 0
b) $\frac{x + 2 }{x – 5}$ < 0
c) $\frac{x - 1}{x + 3}$ > 1
Comments ( 1 )
a)\left[ \begin{array}{l}
x > 3\\
x < 2
\end{array} \right.\\
b)5 > x > – 2\\
c)x < – 3
\end{array}\)
a)DK:x \ne 3\\
\dfrac{{x – 2}}{{x – 3}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 2 > 0\\
x – 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 < 0\\
x – 3 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > 3\\
x < 2
\end{array} \right.\\
b)DK:x \ne 5\\
\dfrac{{x + 2}}{{x – 5}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
x – 5 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
x – 5 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > – 2\\
x < 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x < – 2\\
x > 5
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 5 > x > – 2\\
c)DK:x \ne – 3\\
\dfrac{{x – 1}}{{x + 3}} > 1\\
\to \dfrac{{x – 1 – x – 3}}{{x + 3}} > 0\\
\to \dfrac{{ – 4}}{{x + 3}} > 0\\
\to x + 3 < 0\\
\to x < – 3
\end{array}\)