a) \(\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x-2+3\right)\left(x-2-3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)
Vậy..
b) \(\left(x-2\right)^2-x^2+4=0\)
\(\Leftrightarrow x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Rightarrow x=2\)
a, \(\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy x = 1 hoặc x = -5
b, \(\left(x-2\right)^2-x^2+4=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-4\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy x = 2
a) \(\left(x+2\right)^2-9=0\) \(\Leftrightarrow\) \(\left(x+2\right)^2=9\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\) vậy \(x=1;x=-5\)
b) \(\left(x-2\right)^2-x^2+4=0\) \(\Leftrightarrow\) \(x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow\) \(-4x+8=0\) \(\Leftrightarrow\) \(-4x=-8\Leftrightarrow x=2\) vậy \(x=2\)
Cách 2:
a, \(\left(x+2\right)^2-9=0\)
\(\Leftrightarrow x^2+4x+4-9=0\)
\(\Leftrightarrow x^2+4x-5=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Vậy...
b, \(\left(x-2\right)^2-x^2+4=0\)
\(\Leftrightarrow x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)
Vậy...
a/ \(\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x+2\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
Vậy................
b/ \(\left(x-2\right)^2-x^2+4=0\)
\(\Leftrightarrow x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow-4x+8=0\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)
Vậy.............
a) \(\left(x+2\right)^2=9\)
\(\left(x+2\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
\(a,\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x+2\right)^2-3^2=0\)
\(\Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)\(b,\left(x-2\right)^2-x^2+4=0\)
\(\Leftrightarrow x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow-4x+8=0\Leftrightarrow-4x=-8\Rightarrow x=2\)
a) Ta có :
\(\left(x+2\right)^2-9=0\)
\(\left(x+2\right)^2=0+9\)
\(\left(x+2\right)^2=9\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=3^2\\\left(x+2\right)^2=\left(-3\right)^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)\(\left(TM\right)\)
Vậy \(x\in\left\{1;-5\right\}\) là giá trị cần tìm
b) \(x^2-4x+4-\left(x^2+4\right)=0\)
\(-4x=0\)
=> x = 0
Ta có :
\(\left(x+2\right)^2-3^2=0\)
\(\left(x+2+3\right)\left(x+2-3\right)=0\)
(x+5)(x-1)=0
xét :
\(x+5=0\)=> x=(-5)
x-1=0=> x=1
Vậy x = (-5) hoặc 1
b, \(\left(x-2\right)^2-x^2+4=0\)
=> \(\left(x-2-x\right)\left(x-2+x\right)+4=0\)
=> (-2)(2x-2)+4=0
=> -4x+4+4=0
=> 8-4x=0
=> 4(2-x)=0
=> 2-x=0=> x=2
Vậy x=2
