a) ĐKXĐ: 1 ≥ x ≥ -1
Ta có: VT ≥ 0 = VP
Dấu "=" xảy ra khi và chỉ khi
\(\left\{{}\begin{matrix}\sqrt{1-x^2}=0\\\sqrt{1+x}=0\end{matrix}\right.\)
<=> x = -1 (TM)
b) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
Ta có: VT ≥ 0 = VP
Dấu "=" xảy ra khi và chỉ khi
\(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x^2+4x+4}=0\end{matrix}\right.\)
<=> x = -2 (TM)
c) \(\sqrt{1-x^2}+\sqrt{x+1}=0\)
ĐKXĐ: \(\left\{{}\begin{matrix}1-x^2\ge0\\x+1\ge0\end{matrix}\right.\) \(\Rightarrow\)\(\left\{{}\begin{matrix}1\ge x^2\\x\ge-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge-1\end{matrix}\right.\)
=> -1 \(\le\) x \(\le\) 1
\(\sqrt{1-x^2}+\sqrt{x+1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(1-x\right)\left(1+x\right)}+\sqrt{x+1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(1+x\right)}.\left(\sqrt{1-x}+1\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{1+x}=0\\\sqrt{1-x}=-1\left(voli\right)\end{matrix}\right.\Rightarrow x+1=0\)
=> x = -1 ( thỏa mãn)
d) ĐKXĐ: \(x^2-4\ge0\Rightarrow x^2\ge4\)
\(\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
\(\sqrt{x^2-4}+\sqrt{\left(x+2^2\right)}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2^2\right)}=0\)
\(\Leftrightarrow\)\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+2=0\\\sqrt{x-2}=-\sqrt{x+2}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x-2=x+2\left(voli\right)\end{matrix}\right.\)
Vậy x= -2
