P=2x+y+30x+5y

=(6x5+30x)+(y5+5y)+(4x5+4y5)

≥2.6+2+45.10=22

Vậy GTNN là P = 22 khi x = y = 5

Ta có: \(P=\frac{9x+18y}{xy}+\frac{2x-5y}{12}+2018\)

\(=\frac{9}{y}+\frac{18}{x}+\frac{x}{6}-\frac{5y}{12}+2018\)

\(=\frac{18}{x}+\frac{x}{2}+\frac{9}{y}+\frac{y}{4}-\frac{x}{3}-\frac{2y}{3}+2018\)

\(=\left(\frac{18}{x}+\frac{x}{2}\right)+\left(\frac{9}{y}+\frac{y}{4}\right)-\frac{x+2y}{3}+2018\)

Vì \(x,y>0\Rightarrow\frac{18}{x}>;\frac{x}{2}>0\)

Áp dụng BĐT cô si cho hai số dương ta có:

\(\frac{18}{x}+\frac{x}{2}\ge2\sqrt{\frac{18}{x}.\frac{x}{2}}=6\)

\(\frac{9}{y}+\frac{y}{4}\ge2\sqrt{\frac{9}{y}.\frac{y}{4}}=3\)

Vì \(x+2y\le18\)

\(\Rightarrow\frac{x+2y}{3}\le\frac{18}{3}=6\)

\(\Rightarrow\frac{-x+2y}{3}\ge-6\)

\(\Rightarrow P\ge6+3-6+2018\)

\(\Rightarrow P\ge2021\)

\(\Rightarrow MinP=2021\Leftrightarrow\hept{\begin{cases}\frac{18}{x}=\frac{x}{2}\\\frac{9}{y}=\frac{y}{4}\\x+2y=18\end{cases}}\)và x,y>0

                                     \(\Leftrightarrow\hept{\begin{cases}x=6\\y=6\end{cases}\Rightarrow x=y=6}\)

\(P=\frac{9}{y}+\frac{18}{x}+\frac{x}{6}-\frac{5y}{12}+2018=\left(\frac{18}{x}+\frac{x}{2}\right)+\left(\frac{9}{y}+\frac{y}{4}\right)-\frac{x}{3}-\frac{2y}{3}+2018\)

Lập luận : Áp dụng BTĐ Cô si cho : \(\frac{18}{x};\frac{x}{2}>0\)(với x  > 0):

\(\frac{18}{x}+\frac{x}{2}\ge2\sqrt{\frac{18}{x}.\frac{x}{2}}\Leftrightarrow\frac{18}{x}+\frac{x}{2}\ge6\)

Lập luận tương tự : Áp dụng BĐT Cô si cho : \(\frac{9}{y};\frac{y}{4}>0\)(y > 0 )

\(\frac{9}{y}+\frac{y}{4}\ge2\sqrt{\frac{9}{y}.\frac{y}{4}}\Leftrightarrow\frac{9}{y}+\frac{y}{4}\ge3\)

Và \(\frac{x}{3}-\frac{2y}{3}=\frac{x+2y}{3}\ge\frac{18}{3}\)(Do x + 2y \(\le\)18)

\(\Rightarrow P=\left(\frac{18}{x}+\frac{x}{2}\right)+\left(\frac{9}{y}+\frac{y}{4}\right)-\frac{x}{3}-\frac{2y}{3}+2018\ge6-3-\frac{18}{3}+2018=2021\)

Vậy \(P=2021\)Khi và chỉ khi \(\hept{\begin{cases}\frac{18}{x}=\frac{x}{2};\frac{9}{y}=\frac{y}{4}\\x+2y< 18;x,y>0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=6\end{cases}}\)

\(P=\frac{18}{x}+\frac{9}{y}+\frac{x}{6}-\frac{5y}{12}+2018\)

\(P=\frac{27}{2x}+\frac{3x}{8}+\frac{9}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}\right)-\frac{5}{24}\left(x+2y\right)+2018\)

\(P\ge2\sqrt{\frac{27.3x}{16x}}+\frac{9}{2}.\frac{9}{x+2y}-\frac{5}{24}.18+2018\)

\(P\ge\frac{9}{2}+\frac{9}{2}.\frac{9}{18}-\frac{15}{4}+2018=2021\)

Dấu "=" xảy ra khi \(x=y=6\)

a/ \(\frac{18}{x}+\frac{x}{2}\ge2\sqrt{\frac{18}{x}.\frac{x}{2}}=6\) (đpcm)

Dấu "=" xảy ra khi \(\frac{18}{x}=\frac{x}{2}\Rightarrow x=6\)

b/

\(P=\frac{9}{y}+\frac{18}{x}+\frac{x}{6}-\frac{5y}{12}+2018\)

\(P=\frac{9}{y}+\frac{y}{4}+\frac{18}{x}+\frac{x}{2}-\frac{1}{3}\left(x+2y\right)+2018\)

\(P\ge2\sqrt{\frac{9}{y}.\frac{y}{4}}+2\sqrt{\frac{18}{x}.\frac{x}{2}}-\frac{1}{3}.18+2018\)

\(P\ge2021\)

\(\Rightarrow P_{min}=2021\) khi \(x=y=6\)

Với a>0,b>0a>0,b>0 ta luôn có a+b≥2ab−−√a+b≥2ab

M = x2+y2xy=xy+yx=3xy+(x4y+yx)x2+y2xy=xy+yx=3xy+(x4y+yx)

Ta có: (x4y+yx)≥2x4y⋅yx−−−−−−√=1(x4y+yx)≥2x4y⋅yx=1

Mặt khác: x≥2yx≥2y ⇒3x4y≥32⇒3x4y≥32

Do đó M≥52M≥52 . Dâu ''='' xảy ra khi x=2yx=2y

Vậy giá trị nhỏ nhất của M là 5252 ⇔x=2y