Đáp án: $MinP=12$
Giải thích các bước giải:
$P=\dfrac{a}{\sqrt{a-1}}+\dfrac{b}{\sqrt{b-4}}+\dfrac{c}{\sqrt{c-9}}$
$\rightarrow P=\dfrac{a-1+1}{\sqrt{a-1}}+\dfrac{b-4+4}{\sqrt{b-4}}+\dfrac{c-9+9}{\sqrt{c-9}}$
$\rightarrow P=(\sqrt{a-1}+\dfrac{1}{\sqrt{a-1}})+(\sqrt{b-4}+\dfrac{4}{\sqrt{b-4}})+(\sqrt{c-9}+\dfrac{9}{\sqrt{c-9}})$
$\rightarrow P\ge 2\sqrt{\sqrt{a-1}.\dfrac{1}{\sqrt{a-1}}}+2\sqrt{\sqrt{b-4}.\dfrac{4}{\sqrt{b-4}}}+2\sqrt{\sqrt{c-9}.\dfrac{9}{\sqrt{c-9}}}$
$\rightarrow P\ge 2+2.2+2.3$
$\rightarrow P\ge 12$
Dấu = xảy ra $\rightarrow a=2, b=8, c=18$
