Question

Giải pt

\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)

Answer 1

Vũ Minh Tuấn Lê Thị Thục Hiền @Nk>↑@ Băng Băng 2k6

Answer 2

\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\) () (đk: \(-\sqrt{10}< x< \sqrt{10}\))

<=>\(\left(x+3\right)\sqrt{10-x^2}=x^2-4x+3x-12\)

<=> \(\left(x+3\right)\sqrt{10-x^2}=\left(x-4\right)\left(x+3\right)\)

<=> \(\left(x+3\right)\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

<=> \(\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}-x+4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\left(tm\right)\\\sqrt{10-x^2}=x-4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\\10-x^2=16-8x+x^2\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-3\\0=6-8x+2x^2\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\\x^2-4x+3=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\\x^2-x-3x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-3\left(tm\right)\\x=1\left(ktm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

Vậy pt (*) có nghiệm duy nhất x=-3