\(F=a^3+b^3+ab\left(a+b\right)+2a+b+\frac{3}{a}+\frac{2}{b}\)

\(F=\left(a+b\right)^3-3ab\left(a+b\right)+ab\left(a+b\right)+a+b+a+\frac{1}{a}+\frac{2}{a}+\frac{2}{b}\)

\(F=8-4ab+2+a+\frac{1}{a}+\frac{2}{a}+\frac{2}{b}\)

Ta có: \(\left(a+b\right)^2\ge4ab\Leftrightarrow-4ab\ge-\left(a+b\right)^2=-4\)

\(a+\frac{1}{a}\ge2\sqrt{a.\frac{1}{a}}=2\)

\(\frac{2}{a}+\frac{2}{b}\ge\frac{8}{a+b}=4\)

Suy ra \(F\ge8-4+2+2+4=12\)

Dấu \(=\)xảy ra khi \(a=b=1\).

mn giúp mình

Bạn kiểm tra lại ạ!!

\(F=2x^2-10x+20=2\left(x-\frac{5}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2},\forall x\)

\(\Rightarrow minF=\frac{15}{2}\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

F = 2x² - 10x + 20

= 2( x² - 5x + 25/4 ) + 15/2

= 2( x - 5/2 )² + 15/2 ≥ 15/2 ∀ x

Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2

=> MinF = 15/2 <=> x = 5/2

                                                   Bài giải

\(F=2x^2-10x+20=2\left(x^2-5x+\frac{25}{4}\right)+\frac{15}{2}=2\left(x-\frac{5}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2}\forall x\)

Dấu " = " xảy ra khi : \(2\left(x-\frac{5}{2}\right)^2=0\text{ }\Rightarrow\text{ }x=\frac{5}{2}\)

Vậy \(Min_F=\frac{15}{2}\text{ khi }x=\frac{5}{2}\)

A =  x^2+2x+2x-3 = x^2+4x-3 = (x^2+4x+4)-7 = (x+2)^2-7 >= -7

Dấu "=" xảy ra <=> x+2 = 0 <=> x=-2

Vậy GTNN của A = -7 <=> x=-2

Tk mk nha 

thak pn:D