a + b + 2ab = 24

<=> a+b = 24 - 2ab

<=> (a +b)^2 = (24 - 2ab)^2

<=> a^2 + b^2 + 2ab = 4a^2*b^2 - 96ab + 576

<=> a^2+b^2 = 4a^2*b^2 - 98ab + 576

Q = a^2 + b^2 = 4a^2*b^2 - 98ab + 576

= 4a^2*b^2 - 2*2*a*b*24,5 + 600,25 - 24,25

= (2ab - 24,5)^2 - 24,25

có: (2ab - 24,5)^2 ≥ 0

=> (2ab - 24,5)^2 - 24,25 ≥ -24,25

vậy gtnn của Q = -24,25 = -97/4

a+b=0 => a=(-b)

=>A=a^2+b^2=a^2+(-a)^2=a^2+a^2=2.a^2\(\ge\)2.0=0

Dấu = xảy ra khi a^2=0 =>a=0 =>b=0

Vậy A_min=0 khi và chỉ khi a=b=0

\(a+b\ge2\sqrt{ab}\Leftrightarrow2\sqrt{ab}\le4\Leftrightarrow ab\le4\)

\(P=\left(\dfrac{2}{a^2+b^2}+\dfrac{1}{ab}\right)+\dfrac{2}{ab}+2ab+\dfrac{32}{ab}\\ \Leftrightarrow P=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{2}{ab}+2ab+\dfrac{32}{ab}\\ \Leftrightarrow P\ge2\cdot\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{32}{ab}\cdot2ab}+\dfrac{2}{4}\\ \Leftrightarrow P\ge\dfrac{8}{\left(a+b\right)^2}+2\sqrt{64}+\dfrac{1}{2}\\ \Leftrightarrow P\ge\dfrac{8}{16}+16+\dfrac{1}{2}=17\)

Dấu \("="\Leftrightarrow a=b=2\)

Ta co:\(1\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le\frac{1}{4}\)

Dat \(P=a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\)

\(=a^2+\frac{1}{16a^2}+b^2+\frac{1}{16b^2}+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)

\(=a^2+\frac{1}{16a^2}+b^2+\frac{1}{16b^2}+\frac{15}{16}.\frac{a^2+b^2}{a^2b^2}\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{16}.\frac{2}{ab}\ge1+\frac{15}{16}.\frac{2}{\frac{1}{4}}=\frac{17}{2}\)

Dau '=' xay ra \(a=b=\frac{1}{2}\)

Vay \(P_{min}=\frac{17}{2}\)khi \(a=b=\frac{1}{2}\)