Toán Lớp 10: Định m để (m-3)x – m + 2 = 0 có nghiệm x < 1.
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Toán Lớp 10: Định m để (m-3)x – m + 2 = 0 có nghiệm x < 1.
Comments ( 1 )
\left( {m – 3} \right).x – m + 2 = 0\\
\Leftrightarrow \left( {m – 3} \right).x = m – 2
\end{array}$
m – 3 \ne 0 \Leftrightarrow m \ne 3\\
\Leftrightarrow x = \dfrac{{m – 2}}{{m – 3}}\\
Khi:x < 1\\
\Leftrightarrow \dfrac{{m – 2}}{{m – 3}} < 1\\
\Leftrightarrow \dfrac{{m – 2}}{{m – 3}} – 1 < 0\\
\Leftrightarrow \dfrac{{m – 2 – \left( {m – 3} \right)}}{{m – 3}} < 0\\
\Leftrightarrow \dfrac{{m – 2 – m + 3}}{{m – 3}} < 0\\
\Leftrightarrow \dfrac{1}{{m – 3}} < 0\\
\Leftrightarrow m – 3 < 0\\
\Leftrightarrow m < 3\\
Vậy\,m < 3
\end{array}$