ĐK: \(a\ge0;b\ge5;c\ge0\)
Phương trình tương đương \(a+b+c-2\sqrt{a}-2\sqrt{b-3}-2\sqrt{c}=0\)
\(\Leftrightarrow\left(a-2\sqrt{a}+1\right)+\left(b-3-2\sqrt{b-3}+1\right)+\left(c-2\sqrt{c}+1\right)=0\)\(\Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b-3}-1\right)^2+\left(\sqrt{c}-1\right)^2=0\)
Từ đó ta phải có: \(\left\{{}\begin{matrix}\left(\sqrt{a}-1\right)^2=0\\\left(\sqrt{b-3}-1\right)^2=0\\\left(\sqrt{c}-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=1\end{matrix}\right.\)
\(a+b+c=2\sqrt{a}+2\sqrt{b-3}+2\sqrt{c}\)
\(\Leftrightarrow a+b+c-2\sqrt{a}-2\sqrt{b-3}-2\sqrt{c}=0\)
\(\Leftrightarrow\left(a-2\sqrt{a}+1\right)+\left[\left(b-3\right)-2\sqrt{b-3}+1\right]+\left(c-2\sqrt{c}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b-3}-1\right)^2+\left(\sqrt{c}-1\right)^2=0\)
\(\left\{{}\begin{matrix}\left(\sqrt{a}-1\right)^2=0\\\left(\sqrt{b-3}-1\right)^2=0\\\left(\sqrt{c}-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=1\end{matrix}\right.\)
Vậy \(a=1\) , \(b=4\) , \(c=1\)
