Câu hỏi của Huỳnh Võ Đông Hân

Để pt có nghiệm duy nhất thì: \(-\dfrac{2}{m}\ne\dfrac{1}{1}\Leftrightarrow m\ne-2\)\(\left\{{}\begin{matrix}-2x+y=-3m-1\\mx+y=m^2+m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)y=m^2+m+3+3m+1\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m^2+4m+4}{m+2}\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\left(m+2\right)^2}{m+2}=m+2\\-2x+\left(m+2\right)=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=m+2+3m+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=4m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\x=\dfrac{4m+2}{2}\end{matrix}\right.\) Mà: \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4m+2}{3}>0\\m+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4m>-2\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{2}\\m>-2\end{matrix}\right.\Leftrightarrow m>-\dfrac{1}{2}\)Câu trả lời: Để pt có nghiệm duy nhất thì: \(-\dfrac{2}{m}\ne\dfrac{1}{1}\Leftrightarrow m\ne-2\)\(\left\{{}\begin{matrix}-2x+y=-3m-1\\mx+y=m^2+m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)y=m^2+m+3+3m+1\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m^2+4m+4}{m+2}\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\left(m+2\right)^2}{m+2}=m+2\\-2x+\left(m+2\right)=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=m+2+3m+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=4m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\x=\dfrac{4m+2}{2}\end{matrix}\right.\) Mà: \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4m+2}{3}>0\\m+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4m>-2\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{2}\\m>-2\end{matrix}\right.\Leftrightarrow m>-\dfrac{1}{2}\)

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Huỳnh Võ Đông Hân

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24 tháng 6 - olm

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HT.Phong (9A5)

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24 tháng 6

Để pt có nghiệm duy nhất thì: \(-\dfrac{2}{m}\ne\dfrac{1}{1}\Leftrightarrow m\ne-2\)

\(\left\{{}\begin{matrix}-2x+y=-3m-1\\mx+y=m^2+m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)y=m^2+m+3+3m+1\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m^2+4m+4}{m+2}\\-2x+y=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\left(m+2\right)^2}{m+2}=m+2\\-2x+\left(m+2\right)=-3m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=m+2+3m+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\2x=4m+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=m+2\\x=\dfrac{4m+2}{2}\end{matrix}\right.\)

Mà: \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4m+2}{3}>0\\m+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4m>-2\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{2}\\m>-2\end{matrix}\right.\Leftrightarrow m>-\dfrac{1}{2}\)

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