Question

[The Moon]

GIÚP EM CÂU C VỚI Ạ,EM CẦN GẤPPPPP

Answer 1

[Nguyễn Hoàng Minh]

\(c,\left\{{}\begin{matrix}-4x+ay=1+a\\\left(6+a\right)x+2y=3+b\end{matrix}\right.\)

Để hpt có nghiệm duy nhất \(\Leftrightarrow\dfrac{-4}{6+a}\ne\dfrac{a}{2}\Leftrightarrow a^2+6a+8\ne0\Leftrightarrow\left\{{}\begin{matrix}a\ne-2\\a\ne-4\end{matrix}\right.\)

Để hpt vô nghiệm \(\Leftrightarrow\dfrac{-4}{6+a}=\dfrac{a}{2}\ne\dfrac{1+a}{3+b}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-4}{6+a}=\dfrac{a}{2}\\\dfrac{a}{2}\ne\dfrac{1+a}{3+b}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=-2\\a=-4\end{matrix}\right.\\2+2a\ne3a+ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=-2\\a=-4\end{matrix}\right.\\a\ne2-ab\end{matrix}\right.\)

Để hpt có vô số nghiệm \(\Leftrightarrow\dfrac{-4}{6+a}=\dfrac{a}{2}=\dfrac{1+a}{3+b}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-4}{6+a}=\dfrac{a}{2}\\\dfrac{a}{2}=\dfrac{1+a}{3+b}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=-2\\a=-4\end{matrix}\right.\\2+2a=3a+ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=-2\\a=-4\end{matrix}\right.\\a=2-ab\end{matrix}\right.\)