Ta có :
\(\left(x+2\right)^2=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
\(\left(x+2\right)^2=0\)
\(< =>x+2=0\)
\(< =>x=-2\)
Ta có : \(\left(x+2\right)^2\ge0\forall x\)
Dấu = xảy ra <=> x + 2 = 0
<=> x = -2
Vậy với x = -2 thì \(\left(x+2\right)^2=0\)
(x+2)^2=0
=>x+2=0
=>x=0-2
=>x=2
Vậy x=2
TA CÓ : (x+2)\(^2\)=0\(\Rightarrow\)x\(^2\)+2\(^2\)=0
x\(^2\)=-4
mà x\(^2\)ko có ở dạng âm\(\Rightarrow\)x\(\in\)\(\varnothing\)
\(\left(x+2\right)^2=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy \(x=-2\)thì \(\left(x+2\right)^2=0\)
(x+2)2=0
\(\Leftrightarrow\)x+2=0
\(\Leftrightarrow\)x= -2
KL : .........
\(\left(x+2\right)^2=0\)
\(x+2=0\)
\(x=-2\)