Đk: \(\left\{{}\begin{matrix}a\ge0\\b\ge1\\c\ge2\end{matrix}\right.\)
pt đã cho \(\Leftrightarrow\left(a-2\sqrt{a}+1\right)+\left(b-1-4\sqrt{b-1}+4\right)+\left(c-2-6\sqrt{c-2}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b-1}-2\right)^2+\left(\sqrt{c-2}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{a}-1=0\\\sqrt{b-1}-2=0\\\sqrt{c-2}-3=0\end{matrix}\right.\)
(bạn tự làm tiếp nhé, nhớ ghi kết luận nha)
ĐKXĐ \(a\ge0;b\ge1;c\ge2\)
Ta có (a - 2\(\sqrt{a}\) +1) + [ (b - 1) - 4\(\sqrt{b-1}\) + 4 ] + [(c - 2 ) - 6\(\sqrt{c-2}\) +9 ] =0
<=> (\(\sqrt{a}\) - 1) + (\(\sqrt{b-1}\) - 2 ) + ( \(\sqrt{c-2}\) - 3 ) =0
<=>\(\left[{}\begin{matrix}\sqrt{a}-1=0\\\sqrt{b-1}-1=0\\\sqrt{c-2}-3=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}a=1\\b=5\\c=11\end{matrix}\right.\)
Vậy .........
