ta có : a = 5 - b \(\Rightarrow a+b=5\)

theo bđt cô si ta có : \(\frac{a+b}{2}\ge\sqrt{ab}\)

                  \(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\ge ab\)

                  \(\Leftrightarrow\frac{\left(a+b\right)^2}{4}\ge ab\)

                  \(\Leftrightarrow\frac{25}{4}\ge ab\) ab đạt GTNN khi ab = \(\frac{25}{4}\)

                 ta có   \(\frac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow a+b\ge2\sqrt{ab}\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\)

                  \(P\ge\frac{2}{\sqrt{ab}}\Leftrightarrow P\ge\frac{2}{\sqrt{\frac{25}{4}}}=\frac{4}{5}\)

dấu " = " xảy ra khi P = 4/5

mik làm lụi >_< 

Thay 1 = abc ta có: \(a+b+c=\frac{abc}{a}+\frac{abc}{b}+\frac{abc}{c}\)

<=> a + b + c = bc + ac + ab

<=> (a - ac) + (b - bc) + (c - ab) = 0 

<=> a(1 - c) + b(1 - c) + (c - \(\frac{1}{c}\)) = 0 

<=> ca(1 - c) + cb(1 - c) + (c - 1)(c + 1) = 0 

<=> (1 - c)(ca + cb - c - 1) = 0 

<=> (1 - c)[c(a -1) + (cb - abc)]= 0 

<=> (1 - c)[c(a - 1) + cb(1 - a)]= 0 

<=> (1 - c)(a - 1)(c - cb) = 0

<=> (1 - c)(a - 1)(1 - b).c = 0 <=> a = 1 hoặc b = 1 hoặc c = 1

Vậy.... 

http://olm.vn/hoi-dap/question/179947.html

\(=\sqrt{\sqrt{2}+\sqrt{6}+\sqrt{3}+3}-\sqrt{2+\sqrt{3}}\)

\(=\sqrt{\sqrt{2}\left(1+\sqrt{3}\right)+\sqrt{3}\left(1+\sqrt{3}\right)}-\sqrt{2-\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\sqrt{2-\sqrt{3}}\)

Nếu đề là \(\sqrt{\sqrt{2}+\sqrt{3}+\sqrt{6}+3}-\sqrt{\sqrt{2}+\sqrt{3}}\) thì ok