\(A=\frac{1}{x-y}+\frac{3xy}{x^3-y^3}+\frac{x-y}{x^2+xy+y^2}\)
Điều kiện : \(x-y\ne0\Leftrightarrow x\ne y\)
\(=\frac{1}{x-y}+\frac{x-y}{x^2+xy+y^2}+\frac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x^2+xy+y^2+\left(x-y\right)^2+3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x^2+4xy+y^2+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2x^2+2xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2}{x-y}\)
Ta có : \(\frac{a^4+b^4}{2}\ge\left(\frac{a+b}{2}\right)^4\) ( BĐT cosi )
\(\Rightarrow a^4+b^4\ge2\left(\frac{a+b}{2}\right)^4\)
\(\Rightarrow\left(2x-3\right)^4+\left(2x-5\right)^4=\left(2x-3\right)^4+\left(5-2x\right)^4\)
\(\ge2\left(\frac{2x-3+5-2x}{2}\right)^4=2\)
Dấu " = " xảy ra khi \(2x-3=5-2x\Rightarrow x=2\)
