a) + b)
Câu hỏi của Trần Đức Mạnh - Toán lớp 9 | Học trực tuyến
c) Đk: \(-\sqrt{10}\le x\le\sqrt{10}\)
\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}=\left(x+3\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(N\right)\\\sqrt{10-x^2}=x-4\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow\left\{{}\begin{matrix}10-x^2=x^2-8x+16\\x\ge4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-8x+6=0\\x\ge4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=3\left(L\right)\\x=1\left(L\right)\end{matrix}\right.\\x\ge4\end{matrix}\right.\)
Kl: x=-3
