\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
a, (–31) . (x +7)=0
<=> x +7 = 0
<=> x = -7
Vậy x \(\in\left\{-7\right\}\)
b, (8 – x) . (x + 13) = 0
<=> \(\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\)
Vậy x \(\in\left\{8;-13\right\}\)
c,(x2– 25) . (3– x )=0
<=> (x - 5) (x + 5) (3 - x) = 0
<=> \(\left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\)
Vậy x \(\in\left\{5;-5;3\right\}\)
d, ( x - 3 ) (x2 + 4) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=3\\x^2=-4\end{matrix}\right.\)(vô lý)
Vậy x \(\in\left\{3\right\}\)
