( x2 - 4 )( x + 35 ) = 0
<=> ( x - 2 )( x + 2 )( x + 35 ) = 0
<=> x = ±2 hoặc x = -35
Vậy ...
Ta có: \(\left(x^2-4\right)\left(x+35\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+35=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=4\\x=-35\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-35\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2;-35\right\}\)
(x2 - 4).(x+35) = 0
TH1 :
x2 - 4 = 0
x2 = 0 +4
x2 = 4
x = 4 : 2
x = 2
TH2:
x + 35 = 0
x = 0 + 35
x = 35
Vậy x thuộc {2; 35}
\(\Rightarrow\)x2-4=0 hoặc x+35=0tương đương x=(0-4):2=-2 và x=0-35=-35
Vậy x=-2 hoặc x=35
\(\left(x^2-4\right)\left(x+35\right)=0\)
TH1 : \(x^2=4\Leftrightarrow x=\pm2\)
TH2 : \(x+35=0\Leftrightarrow x=-35\)
àm lại nì:
x2 - 4).(x+35) = 0
TH1 :
x2 - 4 = 0
x2 = 0 +4
x2 = 4
x = 4 : 2
x = ±2
TH2:
x + 35 = 0
x = 0 - 35
x = -35
Vậy x thuộc {±2; -35}
(x2-4).(x+35)=0
\(\Rightarrow\)x2 - 4 = 0 hoặc x + 35 = 0
TH1: x2 - 4 = 0 TH2 : x + 35 = 0
x2 = 0 + 4 x = 0 - 35
x2 = 4 x = -35
x2 = 22
\(\Rightarrow x\in\left\{2;-2\right\}\)
\(\Rightarrow\)\(x\in\left\{-35;-2;2\right\}\)
Đầy đủ;; Phù hợp với lớp 6
Ta có: \(\left(x^2-4\right)\left(x+35\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+35\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x+35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-35\end{matrix}\right.\)
Vậy: S={2;-2;-35}