\(\sqrt{ab}=a+b\)
a^2+2ab+b^2=10a+b
a^2+2(b-5)a+b^2-b=0
a^2+2(b-5)a+(b-5)^2+9b-25=0
(a+(b-5)^2=25-9b
(a+(b-5)^2>=0\(\hept{\begin{cases}25-9b\ge0\Rightarrow b\le3\\25-9b=k^2\Rightarrow b=\left\{0,1\right\}\end{cases}}\)
\(b=0\Rightarrow\left(a-5\right)^2=25\Rightarrow\orbr{\begin{cases}a=0\left(loai\right)\\a=10\end{cases}}\)
\(b=1\Rightarrow\left(a-4\right)^2=16\Rightarrow\orbr{\begin{cases}a=0\left(loai\right)\\a=8\end{cases}}\)
Kết luận:
ab =100
ab=81
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