Sory, bài khi nãy mk làm sai nên xóa. Làm lại:
\(\sqrt{25-x^2}-\sqrt{10-x^2}=3\left(ĐK:x\le\sqrt{10}\right)\)
Đặt: \(\left\{{}\begin{matrix}a=\sqrt{25-x^2}\\b=\sqrt{10-x^2}\end{matrix}\right.\)
Khi đó: \(\left\{{}\begin{matrix}a-b=3\\a^2-b^2=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3+b\\\left(3+b\right)^2-b^2=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3+b\\9+6b+b^2-b^2=15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=b+3\\6b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3+b\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=1\end{matrix}\right.\)
Suy ra: \(\left\{{}\begin{matrix}\sqrt{25-x^2}=4\\\sqrt{10-x^2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-x^2=16\\10-x^2=1\end{matrix}\right.\)
\(\Leftrightarrow x^2=9\Leftrightarrow\left[{}\begin{matrix}x=3\left(n\right)\\x=-3\left(n\right)\end{matrix}\right.\)
\(\sqrt{25-x^2}+\sqrt{10-x^2}=3\left(ĐK:x\le\sqrt{10}\right)\)
Đặt: \(a=\sqrt{25-x^2},b=\sqrt{10-x^2}\)
Khi đó:
\(\left\{{}\begin{matrix}a+b=3\\a^2-b^2=25-x^2-10+x^2=15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\a^2-b^2=15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\a^2-\left(3-a\right)^2=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\a^2-\left(9-6a+a^2\right)=15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\a^2-9+6a-a^2=15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\6a=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\a=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-1\\a=4\end{matrix}\right.\)
Suy ra: \(\left\{{}\begin{matrix}\sqrt{25-x^2}=4\\\sqrt{10-x^2}=-1\left(loai\right)\end{matrix}\right.\) \(\Leftrightarrow vonghiem\)
P/S: mk nghĩ thế, không chắc
