\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)

\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)

\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\Rightarrow\dfrac{y}{x}\ge4\)

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

Đặt \(\dfrac{y}{x}=a\ge4\Rightarrow P=\dfrac{2a^2-2a+1}{a+1}=2a-4+\dfrac{5}{a+1}\)

\(P=\dfrac{a+1}{5}+\dfrac{5}{a+1}+\dfrac{9}{5}.a-\dfrac{21}{5}\ge2\sqrt{\dfrac{5\left(a+1\right)}{5\left(a+1\right)}}+\dfrac{9}{5}.4-\dfrac{21}{5}=5\)

Dấu "=" xảy ra khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

đây là những món quà mà bn sẽ nhận đc: 1: áo quần 2: tiền 3: đc nhiều người yêu quý 4: may mắn cả 5: luôn vui vẻ trong cuộc sống 6: đc crush thích thầm 7: học giỏi 8: trở nên xinh đẹp phật sẽ ban cho bn những điều này nếu cậu gửi tin nhắn này cho 25 người,

Có: \(A=16xy+\dfrac{1}{xy}-15xy\)

Áp dụng bdt Co-si, ta có:

\(16xy+\dfrac{1}{xy}\ge2\sqrt{16xy.\dfrac{1}{xy}}=8\)

Có \(x+y\ge2\sqrt{xy}< =>xy\le\dfrac{1}{4}\)

=> A \(\ge8-15.\dfrac{1}{4}=\dfrac{17}{4}\)

Dấu "=" xảy ra <=> x = y= \(\dfrac{1}{2}\)

\(K=\left(4xy+\dfrac{1}{4xy}\right)+\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{5}{4xy}\)

\(K\ge2\sqrt{\dfrac{4xy}{4xy}}+\dfrac{4}{x^2+y^2+2xy}+\dfrac{5}{\left(x+y\right)^2}\ge2+4+5=11\)

\(K_{min}=11\) khi \(x=y=\dfrac{1}{2}\)

\(\Leftrightarrow6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+20=\dfrac{5\left(x+y\right)\left(xy+3\right)}{xy}\ge\dfrac{5\left(x+y\right)2\sqrt{3xy}}{xy}=10\sqrt{3}\left(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}\right)\)

Đặt \(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=t\ge2\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\)

\(\Rightarrow6\left(t^2-2\right)+20\ge10\sqrt{3}t\)

\(\Rightarrow3t^2-5\sqrt{3}t+4\ge0\)

\(\Rightarrow\left(\sqrt{3}t-1\right)\left(\sqrt{3}t-4\right)\ge0\)

Do \(t\ge2\Rightarrow\sqrt{3}t-1>0\)

\(\Rightarrow\sqrt{3}t-4\ge0\Rightarrow t\ge\dfrac{4}{\sqrt{3}}\)

\(\Rightarrow t^2\ge\dfrac{16}{3}\Rightarrow t^2-2\ge\dfrac{10}{3}\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge\dfrac{10}{3}\) (do \(\dfrac{x}{y}+\dfrac{y}{x}=t^2-2\))

Vậy \(A_{min}=\dfrac{10}{3}\) khi \(\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

khó thế