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

\(VT=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x-y\right)^2}{x^2y^2}+\dfrac{2}{xy}\ge2\sqrt{\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2x^2y^2}}+\dfrac{2}{xy}=\dfrac{2}{\left|xy\right|}+\dfrac{2}{xy}\ge\dfrac{2}{xy}+\dfrac{2}{xy}=\dfrac{4}{xy}\)

\(xy\ne0,x,y\ne1\)

\(A=\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x+y\right)}{x^2y^2+3}\)

\(xét:\dfrac{2\left(x+y\right)}{x^2y^2+3}=\dfrac{2}{x^2y^2+3}\left(1\right)\)

\(\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}\left(2\right)\)

\(xét:\) \(x^4-x-y^4+y=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3-1\right)\)

\(=\left(x-y\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)+xy\left(x+y\right)-1\right]\)

\(=\left(x-y\right)\left(1-3xy+xy-1\right)\)

\(=\left(x-y\right)\left(-2xy\right)=-2xy\left(x-y\right)=2xy\)

\(xét\) \(\left(y^3-1\right)\left(x^3-1\right)=x^3y^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)

\(=x^3y^3-\left(1-3xy\right)+1=x^3y^3+3xy=xy\left(x^2y^2+3\right)\)

\(\Rightarrow\left(2\right)\Leftrightarrow\dfrac{-2\left(x-y\right)}{x^2y^2+3}\)

\(\left(1\right)\left(2\right)\Rightarrow A=\dfrac{2}{x^2y^2+3}-\dfrac{2\left(x-y\right)}{x^2y^2+3}=\dfrac{2-2x+2y}{x^2y^2+3}\ne0\left(đề-sai\right)\)

Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:

\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

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

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

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