Ta có: \(M=\frac{9}{xy}+\frac{17}{x^2+y^2}\) 

\(=\frac{18}{2xy}+\frac{17}{x^2+y^2}\) 

\(=\left(\frac{17}{x^2+y^2}+\frac{17}{2xy}\right)+\frac{1}{2xy}\) 

Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)(x,y>0), ta có: 

\(M\ge\frac{17.4}{\left(x+y\right)^2}+\frac{2}{\left(x+y\right)^2}=\frac{68}{256}+\frac{2}{256}=\frac{35}{128}\)  

Dấu "=" xảy ra khi: \(x=y=8\)

đề sai

ko tin bạn đọc lại đề xem,nó vòng lặp sai mà,cái đầu tiên đó

minA=0

Do \(x-y=\dfrac{x+y}{\sqrt{xy}}>0\Rightarrow x>y\)

Khi đó:

\(\sqrt{xy}\left(x-y\right)=x+y\Rightarrow xy\left(x-y\right)^2=\left(x+y\right)^2\)

\(\Rightarrow xy\left[\left(x+y\right)^2-4xy\right]=\left(x+y\right)^2\)

\(\Rightarrow\left(xy-1\right)\left(x+y\right)^2=4x^2y^2\)

\(\Rightarrow\left(x+y\right)^2=\dfrac{4x^2y^2}{xy-1}\)

Do vế trái dương nên vế phải dương \(\Rightarrow xy-1>0\)

\(\Rightarrow\left(x+y\right)^2=\dfrac{4x^2y^2-4+4}{xy-1}=4xy+4+\dfrac{4}{xy-1}=4\left(xy-1\right)+\dfrac{4}{xy-1}+8\)

\(\ge2\sqrt{4\left(xy-1\right).\dfrac{4}{xy-1}}+8=16\)

\(\Rightarrow x+y\ge4\)

\(P_{min}=4\) khi \(\left(x;y\right)=\left(2+\sqrt{2};2-\sqrt{2}\right)\)

\(P=\dfrac{1}{2021}\left(\dfrac{2021^2}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{2021}.\dfrac{\left(2021+1\right)^2}{x+y}=\dfrac{1}{2021}.\dfrac{2022^2}{\dfrac{2022}{2021}}=2022\)

\(P_{min}=2022\) khi \(\left(x;y\right)=\left(1;\dfrac{1}{2021}\right)\)

Trừ vế cho vế:

\(xy+z-\left(x+yz\right)=1\)

\(\Leftrightarrow x\left(y-1\right)-z\left(y-1\right)=1\)

\(\Leftrightarrow\left(x-z\right)\left(y-1\right)=1\)

Do \(y\) nguyên dương \(\Rightarrow y\ge1\Rightarrow y-1\ge0\Rightarrow x-z>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-z=1\\y-1=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=2\\z=x-1\end{matrix}\right.\)

Thế vào \(x+yz=2020\)

\(\Rightarrow x+2\left(x-1\right)=2020\)

\(\Leftrightarrow3x=2022\Rightarrow x=674\Rightarrow z=673\)

Vậy \(\left(x;y;z\right)=\left(674;673;2\right)\)

nhân M vs 4 đc \(\frac{3x^2+\left(x-2y\right)^2+4xy}{xy}=\frac{3x}{y}+\frac{\left(x-2y\right)^2}{xy}+4\)

x-2y>=0   và x>=2y => 3x/y>=6   => 4M >=10

x/y+y/x=x^2+y^2/xy​                       sử dụng bdt cosi =>x^2+y^2/xy+xy/x^2+y^2>=1

ta có: \(M=\frac{x}{y}+\frac{y}{x}+\frac{xy}{x^2+y^2}=\frac{x^2+y^2}{xy}+\frac{xy}{x^2+y^2}\ge2\sqrt{\frac{x^2+y^2}{xy}\cdot\frac{xy}{x^2+y^2}}=2\cdot\sqrt{1}=2\cdot1=2.\)

(Ở đây mình áp dụng BĐT Cauchy: \(a+b\ge2\sqrt{ab}\)nhé!)

Học tốt! ^3^