a, \(\sqrt{4\left(x+1\right)}=\sqrt{8}\Leftrightarrow4\left(x+1\right)=8\)
\(\Leftrightarrow x+1=2\Leftrightarrow x=1\)
b, \(\sqrt{x^2-4}-\sqrt{x-2}=0\Leftrightarrow\sqrt{\left(x+2\right)\left(x-2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x+2}.\sqrt{x-2}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{x+2}=-1\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=2\)
a , \(\sqrt{4\left(x+1\right)}=\sqrt{8}\)
Bình phương 2 vế ta có :
\(\Leftrightarrow\left[\sqrt{4\left(x+1\right)}\right]^2=\left(\sqrt{8}\right)^2\)
\(\Leftrightarrow4x+4=8\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
b , \(\sqrt{x^2-4}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x^2-4}=\sqrt{x-2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-4=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\)
a) \(\sqrt{4\left(x+1\right)}=\sqrt{8}\)
⇔ 4(x +1) = 8
⇔ x + 1 = 2
⇔ x = 1
b) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)
⇔ \(\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)
⇔ \(\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)
⇔ \(\sqrt{x-2}.\left(\sqrt{x+2}+1\right)=0\)
⇔\(\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}+1=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=-1\end{matrix}\right.\)
⇔ x = 2
