Question

Giari pt

\(\left(x^2-5\right)\left(x+3\right)=0\)

Answer 1

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=5\\x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\\x=-3\end{matrix}\right.\)

Answer 2

\(=>\left[{}\begin{matrix}x^2-5=0\\x+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\pm\sqrt{5}\\x=-3\end{matrix}\right.\)

vậy.....

Answer 3

⇔[x2−5=0x+3=0⇔[x2−5=0x+3=0

⇔[x2=5x=−3⇔[x2=5x=−3

⇔⎡⎢⎣x=−√5x=√5x=−3

 

Answer 4

<=>\(\left[{}\begin{matrix}x^2-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\\x=-3\end{matrix}\right.\)

Vậy S=\(\left\{-3,\sqrt{5},-\sqrt{5}\right\}\)