Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)

\(P=\dfrac{4k^2-2k.3k+9k^2}{4k^2+2k.3k+9k^2}=\dfrac{13k^2-6k^2}{13k^2+6k^2}=\dfrac{7k^2}{19k^2}=\dfrac{7}{19}\)

theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)

A=\(\dfrac{x-y}{x+y}\)

=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)

=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)

vì y>x>0=> A=-1/2

\(S=\dfrac{x^2+y^2+2xy}{x^2+y^2}+\dfrac{x^2+y^2+2xy}{xy}\)

\(=1+\dfrac{2xy}{x^2+y^2}+2+\dfrac{x^2+y^2}{xy}\)

\(=3+\dfrac{2xy}{x^2+y^2}+\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}\)

\(\dfrac{2xy}{x^2+y^2}+\dfrac{x^2+y^2}{2xy}>=2\cdot\sqrt{\dfrac{2xy}{x^2+y^2}\cdot\dfrac{x^2+y^2}{2xy}}=2\)

Dấu = xảy ra khi \(\dfrac{x^2+y^2}{2xy}=\dfrac{2xy}{x^2+y^2}\)

=>x=y

x^2+y^2>=2xy

=>\(\dfrac{x^2+y^2}{2xy}>=1\)

Dấu = xảy ra khi x=y

=>S>=6

Dấu = xảy ra khi x=y

\(\dfrac{x^2+y^2}{xy}=t;x,y>0\Rightarrow t\ge2\) khi x=y

\(A=t+\dfrac{1}{t}\ge2+\dfrac{1}{2}=\dfrac{5}{2}\)

\(A-\dfrac{5}{2}=\left(t-2\right)+\left(\dfrac{1}{t}-\dfrac{1}{2}\right)=\left(t-2\right)-\dfrac{\left(t-2\right)}{2t}=\dfrac{\left(2t-1\right)\left(t-2\right)}{2t}\)

\(t\ge2\Rightarrow\left\{{}\begin{matrix}2t-1>0\\t-2\ge0\\2t>0\end{matrix}\right.\)\(\Rightarrow\dfrac{\left(2t-1\right)\left(t-2\right)}{2t}\ge0\) đẳng thức khi t=2

\(\Rightarrow A-\dfrac{5}{2}\ge0\Rightarrow A\ge\dfrac{5}{2}\)

Vậy GTNN (A) =5/2 khi x=y

Áp dụng BĐT Cauchy-Schwarz dạng Engel có:

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{\left(x+y\right)^2}=6\)

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

áp dụng BDT AM-GM

\(=>x+y\ge2\sqrt{xy}=>1\ge2\sqrt{xy}=>\sqrt{xy}\le\dfrac{1}{2}=>xy\le\dfrac{1}{4}\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\)

\(\ge\dfrac{4}{x^2+2xy+y^2}+\dfrac{1}{2.\dfrac{1}{4}}=\dfrac{4}{\left(x+y\right)^2}+2=4+2=6\)

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

h

đề này sao tính đc. Bn ghi thiếu đề rồi, cái đó là cho trước mà...

Ta có bất đẳng thức phụ: \(xy+yz+xz\le x^2+y^2+z^2\)

\(\Rightarrow xy+yz+xz\le x^2+y^2+z^2\le3\)

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(P=\dfrac{1}{1+xy}+\dfrac{1}{1+xz}+\dfrac{1}{1+yz}\ge\dfrac{\left(1+1+1\right)^2}{1+xy+1+xz+1+yz}\ge\dfrac{\left(1+1+1\right)^2}{1+1+1+3}=\dfrac{9}{6}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi: \(x=y=z=1\)