a)\(x^2=0\\ \Leftrightarrow x=0\)
vậy...
b)\(x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
vậy...
c)\(x^2=2\\ \Rightarrow x^2=\left(\pm\sqrt{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
vậy...
d)\(x^2=6\left(x>0\right)\\ \Rightarrow x^2=\left(\pm\sqrt{6}\right)^2\\ màx>0\\ \Rightarrow x=\sqrt{6}\)
vậy...
e)\(x^2=7\left(x< 0\right)\)
\(wtf\) ????? thông minh đấy \(x^2\ge0\) mà điều kiện lại là x < 0 ??? :D
rỗng r
f) \(\left(x+1\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
vậy....
g)\(\left(x-2\right)^2=2\\ \Rightarrow\left(x-2\right)^2=\left(\pm\sqrt{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=\sqrt{2}\\x-2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+2\\x=-\sqrt{2}+2\end{matrix}\right.\)
tự tính :D
vậy..
h)\(\left(x+\sqrt{3}\right)^2=5\\ \Leftrightarrow\left(x+\sqrt{3}\right)^2=\left(\pm\sqrt{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\sqrt{3}=\sqrt{5}\\x+\sqrt{3}=-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\\x=\end{matrix}\right.\)
tự tính lười lắm
