\(\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)
Ta có \(\left\{{}\begin{matrix}\sqrt{\left(x-2\right)^2+1}\ge\sqrt{1}=1\\\sqrt{\left(x-2\right)^2+4}\ge\sqrt{4}=2\\\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{matrix}\right.\) \(\Rightarrow VT\ge1+2+\sqrt{5}=3+\sqrt{5}\)
Dấu "=" xảy ra khi và chỉ khi \(x=2\)
b/ Tương tự:
\(\sqrt{3-\left(x-1\right)^2}+\sqrt{1-\left(x+3\right)^2}=1+\sqrt{3}\)
\(\left\{{}\begin{matrix}\sqrt{3-\left(x-1\right)^2}\le\sqrt{3}\\\sqrt{1-\left(x+3\right)^2}\le\sqrt{1}=1\end{matrix}\right.\) \(\Rightarrow VT\le1+\sqrt{3}\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow x=\varnothing\)
Phương trình vô nghiệm
